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Quantitative Analysis

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Quantitative Analysis



Profoh-sor of Chemistry in Vassar College

! . ." I

Philadelphia .TOE BLAKIS-TON. COMPANY .Woronlo

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This hook is fully protected by copyright, and no

part of i/, with the exception of short quotations

for review, may be reproduced without the written

consent of the publisher


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is always important to get off to a good start in any undertaking.

Particularly in a first course of study in quantitative chemistry is it

desirable that the initial experimental work be of a nature that will stress

the necessity of rigor without overwhelming the student with a deluge of

minutiae. As a beginning I prefer the determination of the sensitivity of

the balance. Among the advantages are these: here the paramount im-

portance of precision is emphasized immediately; the contribution of a

large number of simple and relevant factors to a single, final result is

demonstrated; and the fidelity of a delicate mechanism is manifested.

Furthermore, here one gets the feel of the balance; as a result, in subse-

quent experiments there is a command of confidence in the ability to

deliver accurate results. If this is followed by the calibration of a volu-

metric instrument where weighings to the nearest milligram will satisfy,

then sufficient familiarity with the balance is attained to meet the more

fastidious demands of weighing to a constant weight in later work. For

such reasons Chapter 2 on general operations and equipment is followed b>

a discussion of accuracy, precision, errors and significant figures, and then

by the use of the balance and instructions in volumetric calibrations.

Volumetric principles and determinations precede those for gravimetric

work, though for courses in which the custom is to reverse this order the*

first four chapters may be followed by Chapters 13, 14 and 15. In each

subdivision of the subject the experimental part immediately follows the

chapter or chapters dealing with the theoretical principles.

The theory and experimental work included in this volume should

require a year's time. By choosing typical procedures from the sections

dealing with volumetric and gravimetric analysis and omitting the more

specialized experiments of the last several chapters, the material for a one-

semester course may be selected. In Chapter 7 a treatment more exten-

sive than customary is given for calculating the data for titration curves.

Lack of time in one-semester courses may necessitate omission of the

material printed in extract type. However, when time allows, a study of

conditions, unusual as compared with more remote points, prevailing in

the majority of titrations at the beginning of the neutralization and im-

mediately around the stoichiometric point is worth the added attention.

Though not greatly altering the course of the curves, these detailed cal-

culations furnish an excellent illustration of the obligation of thorough-

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ness, and also provide further opportunity for the observation of commonion and hydrolytic effects.

It seems to me that a course in elementary quantitative analysis should

not presume a previous knowledge of chemistry other than that acquired

in a sound course in general chemistry and qualitative analysis. In pre-

paring the text I have kept this in mind. Occasionally, as in presenting the

theory of indicators or in taking up the efficiency of desiccants, J find it

desirable to include concepts which the student will explore more fully in

later studies; however, in such cases the presentation has been kept on

a simple basis. On the other hand, it hardly seems necessary to review such

foundation laws as that of combining weights or the law of mass action,

which have already been studied and applied.

During the past decade and more I have used for my own classes in

quantitative chemistry the excellent texts of Blasdale; Booth and Damcr-

ell; Fales and Kenny; Kanning; Kolthoff and Sandoll; Piorce and Hauu-

isch; Rieman, INeuss and Nairnan; Talbot (Hamilton and Simpson); and

Willard and Furman. My own thinking has been influenced in large

measure by the pedagogy of these authors and my indebtedness to them is

unqualified. I am grateful to Dr. Mary 0. Ilillis for contributing manyhours to methods of presenting various topics and for her valuable

criticism of many sections of the manuscript. My thanks are also due to

Mrs. Charles M. Kuhbach who not only took great pains in typing the

manuscript but also made many helpful editorial suggestions. The advice

of The Blakiston Company was frequently sought and generously given;

particular acknowledgment is due to the Editor, Dr. James B. Lackey, for

his interest and attention. I shall hoartily welcome any notice of errors

and of ideas for improvement of the text.


Poughkeepsie, New York

March 1919

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Potassium Permanganate Processes .... 187Ceric Processes .... . 199Potassium Diehromate Processes 201










Care of Platinum Ware 327References 329

Tables 331

Logarithms 338


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(a) With Potassium Biphthalate . 136

(b) With Sulfamic Acid '. 136

(c) With Benzoic Acid 137












(a) With Sodium Oxalate 189

(b) With Arsenious Qxide 189

(c) With Iron Wire 190

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Zimmermann-Reinhardt Reduction ............. 191

Jones Reductor ...................... 192

Silver Reductor ...................... 194

ANALYSIS OF PYROLUSITE................... 195

DETERMINATION OF CALCIUM ................. 196





(a) With Arsenious Oxide ...... 200

(b) With Iron Wire ......... .201















(a) With Potassium Dirhromato . .212(b) With Potassium Bromate, lodato or Bi-iodate . 2'i3

(c) With Metallic Copper. ... 214










(a) Phosphomolybdate Method...... ' ..... 256

(b) Short Method ........... . . . 259

DETERMINATION OF IRON................. 260

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(a) Moisture and Loss on Ignition 263

(by Silica. 264

(t) Combined Oxides . . . .... 266

(T# Calcium . 267

(e) Magnesium 269

(f) Carbon dioxide . .... 270



















(a) Tin 312

(b) Lead 313

(c) Copper 314

(d) Iron . . 314

(e) Zinc 314



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T,HE ULTIMATE goal of every scientific journey has the peculiarity,

which at the same time is its fascination, of belonging to the sphere of the

unknown. At the furthest borders of encompassed knowledge, of that

area which has been probed and charted by the patient and often in-

genious labors of man, one stands at the pinnacle of recorded humanachievement only to see beyond far vaster regions still wrapped in

obscurity and mystery. It is this realm of the unknown with its promise

of discovery, its invitation to behold that which never before has been

revealed to man, that marks the aim of scientific enterprise, as goal, as

challenge and as final reward for the dedication and the effort required of

him who aspires to become a scientist.

There is always a compelling persuasion in discovery. This is as it

should be; otherwise the accumulation of knowledge loses something of its

zest. Much has been said about the exhilaration of reaching the horizon

from which one may look out upon and then explore the unknown. But

how little has been told us of the joys of the journey to these borders.

Though already well charted, the route is no parkway for effortless travel.

We look about and see such a threatening expanse over which we must

pass before we emerge upon higher ground that sometimes we are all but

dismayed. Just how do we get there? Not by a superficial survey, byrandom wandering, nor by unorganized thrashing at the underbrush;

one cuts through to the broad avenues which lead to the boundless

stretch beyond by an orderly orientation, through a firm resolution to take

each barricade one by one, and in a constant vigilance for the unforeseen.

The rewards of this preliminary ascent which we are about to begin are

rich and the labor involved brings its own satisfaction. For while we are

yet striving toward the goal we know the joy of progress and the gratifica-

tion which comes with a growing power of perception.

Once the terrain of higher ground has been reached we shall find manyavenues leading on to the horizon. When these have been gained the

journey will be more swift. It makes little difference which of these high-

ways later we may choose; the habits of perseverance, exactness, orderli-

ness and impartiality of judgment, which we learn as we clear these

simpler approaching pathways, will be found to be the means also of

surmounting obstacles as yet remote. And nowhere more than in quanti-

tative analysis does one acquire discipline in avoiding detours; nowhere

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does one learn better the priceless value of assurance gained after each

advance through what, a few miles back, might have appeared as a

formidable tract; nowhere else does one more quickly sense with what sure

success careful thought and skilful technique sweep away entanglements.

Here then is the opportunity for joy: the joy of finding the tasks which

lie ahead worthy of the effort, the joy of constructing a pathway cleanly

penetrating what was once strange land, and the joy of commanding new

knowledge and new skills. Hero, at last, we shall find the satisfaction that,

having crossed this barrier, we are better seasoned for the further travels.

And as each such barrier is met and crossed the disclosure becomes ever

clearer that the ultimate goal boyond is no mirage, but is reality itself,

worthy of all the labor and all the joy.


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Chapter 1


V/UANTITATIVE analysis, as the term implies, is concerned with the

determination of the amount of one or more of the constituents presentin a substance. If little or nothing is known about the composition of

the substance under study it will be necessary to make a qualitative

analysis in order to learn what constituents make up the material. The

quantitative analysis may be of a gravimetric nature. This involves the

separation and weighing of the constituent sought, as such or as one of

its compounds of definite composition. The separation may be effected

sometimes by extraction (p. 13), by electrolytic deposition (Chapter 18),

by evolution as a gas (p. 270), but most often by the precipitation of a

compound containing the constituent and bearing a known and invari-

able mathematical relation to it (Chapter 15). The analysis may be of a

volumetric nature. This involves the measurement of the volume of a solu-

tion of known concentration which is required to react with the sample.The completion of the desired reaction is marked by such means as

the color change of indicator dyes (pp. 100, 184 and Chapter 8) or by the

sudden change in some such property as an electrode potential (Chapter

17). The analysis may be accomplished by special methods which lie

outside the above two classifications; these depend upon the determi-

nation of some specific property of the constituent sought which varies

in proportion to the quantity present. The amount of alcohol in a solu-

tion, for example, may be determined by taking the specific gravity of

the measured distillate from a definite volume of an aqueous solution.

Again the index of refraction of many binary mixtures (solutions) will

reveal the composition. Further, the intensity of color imparted by a

colored component to a solution may be made to indicate the quantity

present (Chapter 19). Other methods which make use of optical proper-ties are: those depending upon the rotation of a plane of polarized light

by a solution of an optically active compound; those in which ihe ab-

sorption or the scattering of light by a suspension is measured; and

spectrographic methods in which the intensity of the lines in the emission

spectrum is a measure of the quantity of the constituent.

Chemical analysis is important from both a practical and a theoretical

point of view. Chemical industry depends upon analysis for guidance in

the purchase of raw materials, as a step by step check on the efficiencyl

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of manufacturing processes and for tests of the final products before

marketing. In theoretical chemistry any hypothesis must undergo ex-

haustive experimental proof before it may be regarded as a generalized

statement of fact, and in such tests it is almost always necessary to

determine the identity and composition of substances resulting from the

study. The practice of analysis therefore may be confined on the one

hand to that of the specialized commercial chemist, or it may be an

adjunct to the investigations of the physical chemist, the organic chemist,

the biochemist and the like. In any event, although there is more of the

routine in his work and though his tools as a rule are more common than

those of the research chemist, the accountability of the analyst is strict;

to meet his responsibilities it is imperative not only that he possess a

high degree of technical skill but that he shall become familiar with the

chemical and physical characteristics and the behavior of a wide variety

of substances. As he learns more about the laws under which chemical

reactions operate he also grows in self-reliance and resourcefulness. Manyof the laws have already been studied in general chemistry and qualitative

analysis. Among those which find wide application in quantitative anal-

ysis are the theory of ionization and the law of mass action and chemical

equilibrium. Based upon these concepts are the principles of the commonion effect, solubility product, hydrolytic and buffer actions and electrode

potentials. Since an elementary understanding of these and other laws

(combining weights, etc.) has been achieved previously, we shall proceedto apply them in the study of quantitative analysis without repeating in

detail their postulates.

Hardly any of the calculations of quantitative chemistry is compli-

cated, being almost wholly confined to arithmetic and algebra, and

seldom calling for anything more involved than a quadratic equation.

However, there is some divergence of practice regarding the unit of the

chemical equivalent. The gram equivalent weight of an element or com-

pound is defined as the weight in grams which reacts with or is displaced

by 1.008 g. of hydrogen or its equivalent. The term gram milliequivalent

weight is often employed in calculations; it is the gram equivalent weightdivided by 1000. The chief reasons why the latter frequently is used is

that it bears the same relation to the former that the milliliter bears to

the liter, and by using milliequivalents one avoids fractions as a rule.

Thus, for example, 0.05845 g. of sodium chloride (eq. wt. = 58.45)

represents 0.001 gram equivalent weight of the salt, or 1 gram milli-

equivalent weight. Here the use of the milliequivalent weight can claim

no greater advantage than the use of pennies rather than dollars could

claim in pricing a commodity. (If an article costs 0.50 dollar we maysay that it is valued at half a dollar or at 50 cents.) It is true that the

number of milliliters of a 1 TV solution is numerically identical with the

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gram milliequivalent weights of solute present: 50 ml. of 1 N solution

contains 50 gram milliequivalent weights; however, it is equally obvious

that 50 ml. of 1 N solution contains 59iooo or 0.05 gram equivalent

weight. Furthermore, in dealing with standard solutions the concentra-

tion is not often exactly 1.000 TV, and the conversion of A nil. ofX normal

solution to its equivalent of B ml. of 1.000 N solution, as sometimes is

done to establish the number of milliequivalent weights, involves just

one extra step. Many analysts therefore prefer to work always on the

basis of the larger unit, the gram equivalent weight. It is somewhat

surprising that there should be any predilection for one system over the

other since cither is quite simple in application. In the calculations

throughout this book both have been used: unless the student already

has developed a choice as a habit in the matter it might be well to em-

ploy one and then the other until a personal preference is felt.

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Chapter 2


D,"URINC, the first period in the study of quantitative analysis the

student is usually assigned to a desk and a balance. In the compartment

of the desk the most commonly used pieces of apparatus will be found,

many of which have been encountered in previous courses of study,, but

some of which will be unfamiliar. It may be necessary to construct such

apparatus as wash bottles. Everyone should learn as soon as possible

the elementary manipulations involved in bending glass and the like,

for the facility of using the hands with dexterity is nowhere more im-

portant than in the quantitative laboratory. Equally important is the

necessity of absolute cleanliness. AH apparatus, reagent bottles and the

balance must be immaculate. The satisfaction of carrying out an analysis

with accurate results, and the confidence which one has in those results,

depend not only upon a thorough understanding of the steps taken

throughout the analysis and upon the proper interpretation of the

measurements made, but also upon the knowledge that no avoidable

error has been allowed to mar those results. This means that every step

must be taken with proper attention given to clean apparatus, pure

reagents and careful technique. Usually the operations are most quickly

arid effectively accomplished by proved methods. Some of these will be

mentioned in this chapter. There are scores of places in every analysis

where errors may creep in; there is only one right way to carry out a

piece of work namely, to see that every step taken is as free from fault

as possible.

Even this is not enough. The best of work is of no permanent value

unless the right kind of record is made. Entries in a bound notebook

should be made in ink, and they should follow a concise, orderly outline

suggested by the instructor so that at a glance one may see the relation

of all the data to each other. Erasures should never be made in the note-

book; if a mistake has been made it should be crossed out neatly and

followed by a new, correct record of the observation or calculation. It is

customary to place on the left-hand page of the notebook all calculations

involved in the determination and such details as equilibrium points of

the balance, and to place on the right-hand page the results of weighings,

buret readings, calculations, etc. The material on the right-hand page

should always be kept up to date; that is, as soon as a calculation has


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been made on the left-hand page the result should be transferred at

once to the right-hand page. All computations should be made duringthe laboratory period and directly in the notebook rather than on odd

scraps of paper or on filter paper.

A great deal of time may be saved by keeping busy during the entire

laboratory period. For example, while a precipitate is being filtered it is

possible to weigh a crucible, or during an ignition to prepare a solution

for later use; and almost always there are calculations that may be madewhile some necessarily slow operation is being carried out. Such time-

saving expedients are anticipated by acquiring a thorough understandingof each procedure before starting it.

It should be borne in mind that other people are working in the

laboratory and sharing many of the reagents. It therefore is the re-

sponsibility of each student to cooperate by exercising care in handling

reagents, refilling reagent bottles and replacing all common laboratory

property at the right location. A substance removed from its container

in greater amount than needed should not be returned to the container

since contamination might result.


Samples. As a rule samples dispensed in the quantitative laboratoryfor analysis already have been properly prepared. That is to say theyare typical of the complete lot of the material from which they were

taken. Moreover, if solid, the material has been ground to proper fine-

ness and often it has been dried. This saves a great deal of students'

time arid permits a greater number of experiments to be performed than

otherwise would be possible. However, unless the beginner has some

experience in preparing a sample, it is likely that he will fail to appreci-ate the cautions necessary in this preliminary step. He should, therefore,

give careful attention to the following considerations.

A sample portion, whether it be, for example, from a mineral deposit,a carload lot of solid material, a tank or vat of liquid or a large body of

water, almost always should be taken in such a manner as to be as

nearly representative of the whole as possible. Thus samples are to be

distinguished from specimens, which are usually gathered to illustrate

particular features and therefore are not typical of the whole. Samplingis not a casual task and great care and judgment are required to secure a

sample for analysis. The chemist should secure his own samples when

possible, or supervise their collection; if neither is possible his reportshould state specifically that the analysis is based on the material as

delivered to him.

The method of taking a sample will depend upon the particular

conditions. The field geologist, for example, is more interested in the

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richness of a given ore not yet mined than he is in the carloads of ore

already taken from the mine. On the other hand, the chemist more often

is called upon to determine the analysis of the ore which, already taken

from the earth, is being purchased by the processor. But in any event

it is imperative that utmost care be taken to secure 3 workable quantityof material which will yield the information wanted. In analyzing a lake

of water information about the dissolved oxygen at various depths maybe desired; in that case the sampling is a simple matter (given the proper

apparatus) of taking the water at the several depths. The same approachis made in taking muck and hurnus samples. Most commonly, however,it is the sample characteristic of the entire lot which is desired, and since

materials like coal, ores and even alloys are not homogeneous, it oten is

necessary to take a gross sample of perhaps 500 to 1000 pounds bycombining "grab" samples taken at random from all parts of the entire

mass. As a rule the number of portions should be not less than 20 andthe total weight of the gross sample should be at least a hundredth of

the entire lot. When the gross sample has been gathered, since it too

will not be homogeneous, it must be reduced in particle size and mixed.

This is often done by mechanical jaw crushers or a steel hammer and

plate, the latter being preferred when possible since it is less likely to

introduce foreign material into the sample. After a given operation has

reduced the particle size the mixing may be accomplished as follows:

The entire quantity is shoveled into a cone, the cone is flattened and the

flat, circular mass is quartered and two diametrically opposite quartersare made into a second cone. After further crushing the process is re-

peated as many times as necessary to yield a proper quantity and particle

size, say about 25 to 1000 g. of 10-mesh particles. It may be necessary,if the material is difficult to dissolve, or if a fusion with a flux is to precedesolution, to reduce to a still smaller size. The ball mill is the most con-

venient apparatus for grinding1 to an extreme state of subdivision, but

the agate mortar and pestle usually suffices if the quantity of material is

not too large. The small quantities which can be ground at one timein the mortar are combined after they have been pulverized and are

thoroughly mixed by placing upon a glazed paper and lifting the corners,

one after the other, thus causing the powdered particles to roll over

themselves ("tabling"). Finally the fine sample is placed in a bottle

and tightly stoppered.

Weighing the Sample for Analysis. The accurate weighing of the

sample involves precautions which call for considerable explanation.

1Grinding should not be carried out for an unnecessarily long time because foreign

substances may be added to the sample from the mill or mortar. Furthermore, exces-

sive grinding may cause the sample to lose water of hydration or otherwise cause a

change in composition, particularly from oxidation.

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The sample may be weighed either directly or by difference. These

points are discussed in the chapter on the balance (p. 39ff).

Solution of the Sample. Whether the analysis is based upon the

precipitation and subsequent weighing of an insoluble compound of the

constituent sought, or upon the interaction of definite volumes of solu-

tions of the unknown and certain reagents, or upon attributes such

as intensity of color, it is first of all necessary to bring the sample into

solution. The simplest means of accomplishing this is to dissolve the

sample in water. However, many substances are practically insoluble in

water. In such cases it frequently happens that treatment with a mineral

acid will suffice. Hydrochloric acid of a concentration of about 1 : 1 with

water should be tried first. Hydrochloric acid is a reducing acid but if

the sample itself is an oxidizing agent the action may be hastened byadding, along with the acid, a reducing agent like sodium bisulfide.

Should the sample contain reducing substances, hydrochloric acid plus

potassium chlorate will be more effective. In the case of metals above

hydrogen in the electromotive series, hydrochloric acid alone is sufficient

for solution; for other metals and for many alloys nitric acid will be best,

or aqua regia may be necessary. If the latter treatment fails to dissolve

the sample it may be treated with perchloric acid 2 or with hydrofluoricacid and nitric acid in a 2:1 ratio. The latter must be carried out in a

platinum crucible and the excess hydrofluoric acid removed by adding2 or 3 ml. of sulfuric acid and evaporating until sulfur trioxide fumes

begin to come off.

It is not necessary to dissolve the entire sample, but only to dissolve

the constituent that is to be determined. For example, hot 1:1 sulfuric

acid often will dissolve the iron of an iron ore, leaving a sandy colored

residue of silica and aluminum silicate. There is no need to bring this

residue into solution unless a complete analysis is desired. The color of

the residue is usually a satisfactory indication as to whether or not all

of the essential constituent has been dissolved, since iron ores are red,brown or black; copper ores are blue, green or black, etc.

Fusions. When the sample resists solution by the above methods it is

necessary to fuse the sample with a suitable flux. The purpose of theflux is to convert the sample into compounds which are soluble either in

water or in acids. Just what flux is best to use depends upon the nature

of the material. Silicates, for example, should be fused with an alkaline

flux of sodium carbonate which converts nonmetals into sodium salts

and metallic constituents into oxides soluble in acids. On the other hand,

many oxide ores, like those of iron and aluminum, require, for completeanalysis, an acid flux such as potassium acid sulfate which changes the

2 Hot concentrated perchloric acid in contact with organic matter should beavoided since such a combination may cause dangerous explosions.

Page 26: quimica inorganica cuantitativa


oxides into soluble sulfates. Oxidizing fluxes, consisting of mixtures of

sodium carbonate and either potassium chlorate, sodium peroxide or

potassium nitrate, should be employed with samples of arsenic ores;

reducing fluxes of sodium thiosulfate or of sulfur mixed with sodium

carbonate are useful in separating many of the metallic sulfides. As a

rule the alkaline fusions should be carried out in platinum crucibles, while

nickel crucibles may be used with an acid flux. Silver or iron crucibles are

recommended for the oxidixing fluxes. When a platinum crucible is

thought proper, consult the instructor before actually starting the fusion.

In any case, the actual fusion is preceded by careful preparation of the

sample and flux. The former must be ground to a fine state of subdivision

and thoroughly mixed with the flux. A small quantity of the flux alone is

put in the crucible, melted and, through a swirling motion, distributed

over the bottom and sides of the crucible to three-quarters of its height.

The mixed sample and flux are then added, a small additional amount of

flux is placed on top of the mix and the crucible then is heated, slowly at

first, until fused. After 30 minutes to an hour, the heat is removed and

the crucible is rotated while cooling in such a manner that the cake solidi-

fies as a crust over the inside wall of the crucible instead of as a small but-

ton in the bottom. When cool, the crucible and contents are placed in a

beaker or casserole, covered with water, or with acid if necessary, and

warmed until complete solution results.

Evaporation. It is frequently necessary in the course of an analysis to

evaporate a liquid in order to reduce its volume, or even to carry the

process to dryness. Evaporations are time-consuming and there is not

much that may be done to hasten the process. Shallow vessels obviously

will give a larger liquid surface and increase the rate somewhat. If the

evaporation is to be to dryness it can be carried out overnight. Care

should be taken in choosing the type of container to be used. Glass, and

to a lesser extent porcelain, is attacked by alkaline solutions, especially

as the concentration increases. If highest accuracy is demanded, platinum

evaporation dishes are to be preferred. Even with platinum high con-

centrations of alkaline solutions must be avoided; such solutions should be

made faintly acid before evaporating.

Evaporations should be made on a steam bath or on an electric hot

plate at a low temperature. The solution should not be allowed to boil

since both spattering and bumping may occur. To prevent dust from fall-

ing into the dish it should be covered with a watch glass resting on a glass

triangle or with the special ribbed watch glass now available which allows

for the escape of the vapor.

Digestion. When a precipitate is first formed from a solution its

state of fineness is so great that often it will pass through the pores of a

filter paper. This difficulty is usually prevented, however, if the particles

Page 27: quimica inorganica cuantitativa


of the precipitate are coagulated or if they may be made to assume a

larger particle size. The latter condition may be brought about by heatingthe precipitate and the mother liquor over a steam bath or hot plate for

an hour or two, or overnight. The solubility of salts is slightly greater

when the solid is extremely finely divided; it follows that a solution which

is saturated with respect to the finely divided particles is, at the same

time, supersaturated with respect to the larger crystals. As a result,

during digestion precipitation will continue upon the coarser particles

at the expense of the finest particles and the latter will eventually dis-

solve. The final result then will be not only the elimination of the finest

crystals, and thus a more easily filtrable precipitate, but actually a small

decrease in the solubility of the precipitate.

Transfer of Liquids. One of the commonest sources of error for the

beginner in the quantitative laboratory results from loss of liquid whenit is transferred from one container to another, or to a filter. Yet it is

easy to avoid this error simply by developing at once the invariable habit

of pouring a liquid down a stirring rod held vertically against the lip of

the vessel. Tn this manner the drop which otherwise would flow from the

lip down the outside of the beaker is retained by the rod. One should

never swing the rod, with its drop or two of liquid clinging to it, between

two vessels. Instead, the rod should always be either in or above one of

the vessels (or the filter). The final step in the transfer of liquids should

be the rinsing of the original container with three successive portions of

the solvent in order to avoid losing the film of solution adhering to the

walls of the container. Note that three small rinse portions are better

than one larger portion; the principle upon which this statement is based

is the same as that which will be discussed later in connection with the

use of many small portions in washing precipitates.

Filtering Precipitates. Before a precipitate is weighed it must be

freed from impurities. To accomplish this it must be filtered and then

washed until both it and the filter are clean. The precipitate is filtered

through a filter paper if it is a compound which either will withstand sub-

sequent ignition without changing its composition or is converted com-

pletely into a compound of definitely known composition. Filter papers

for quantitative work have been treated with hydrochloric and hydro-

fluoric acids so that they will be Cashless" after ignition. The weight of

the ash is marked on the box containing the papers and should be less

than 0.1 mg. Filter papers may be purchased which meet various needs.

Some will retain finer precipitates than others, but usually the finer the

pores of the paper the slower will be the filtration. The paper selected for

a given filtration should be chosen with due consideration of the nature of

the precipitate.

If the substance filtered is one like silver chloride which cannot be

Page 28: quimica inorganica cuantitativa


ignited without decomposition, it must be filtered through a medium

other than paper. The Gooch crucible, made either of platinum or of

porcelain, is the crucible" most frequently used in such instances. It is

much like an ordinary'crucible exclpFltmtrtiie bottom is perforated. A

"soup" of asbestos suspended in water is poured into the crucible and

suction is applied (Fig. 1) so that a mat about as thick as a dime is packed

down upon the perforated bottom. Then a perforated disk is placed upon

the mat and more asbestos is poured in upon it. After suction again is

applied the result is a mat about twice the thickness

of a dime with the disk sandwiched in between the

two layers. Under moderate suction water should drip

through the crucible; if it streams through, the asbes-

tos is too thin and should be replaced with a thicker


Porous porcelain crucibles and sintered glass cru-

cibles are often used instead of Gooch crucibles. In

both cases the bottom of the crucible is sufficiently

porous to function as a filter. Suction is applied in the

same manner as with a Gooch crucible.

Washing Precipitates. Since the precipitate is

always formed in the presence of the ions of the super-

natant liquid it is necessary that thorough washing

follow the filtration. The time expended in washing is

shortened if the precipitate is first washed by decanta-

tion. After the precipitate has settled in the beaker,

almost all of the liquid is passed through the filter

while the precipitate largely remains in the beaker

(Fig. 2). Then wash solution (or water) is added and agitated with the

precipitate. The precipitate is allowed to settle and the liquid is again

poured off into the filter. Wash solution is added to the beaker a second

time and stirred, the precipitate allowed to settle and the new superna-

tant liquid decanted into the filter as before. After this operation has

been repeated three or four times the entire precipitate is transferred to

the filter in a manner illustrated in Fig. 3, that is, by placing the stirring

rod across the top of the beaker and holding over the filter as illustrated,

after which a stream of water or wash solution is directed into the beaker

so that the entire inside of the beaker is washed free of precipitate. If any

particles of precipitate adhere to the beaker it may be necessary to

detach them with a policeman (a glass rod equipped with a rubber tip).

In this case both beaker and policeman must be flushed again with water

which goes thence into the filter.

Often it happens that a filter, while showing no leakage during the

filtration, will not retain the finest particles of the precipitate after the

FIG. 1. Goochcrucible and suc-

tion flask.

Page 29: quimica inorganica cuantitativa


FIG. 2. (Left). Decanting a liquid from beaker to filter.

FIG. 3. (Right). Flushing precipitate from beaker to filter.

washing has been continued to the point where the precipitate is practi-

cally clean. This happens because, so long as the precipitate was not clean,

there were sufficient ions present to prevent any particles from approach-

ing the colloidal state, but just as soon as the precipitate is virtually clean

there are no longer enough ions to cause flocculation; as a result some

precipitate may leak through the filter. This peptization can be prevented

by washing with a dilute solution of a suitable electrolyte instead of with

pure water. The electrolyte used must be one that may be removed byvolatilization during the ignition of the filter paper. Thus dilute acids,

ammonium hydroxide or ammonium salts are commonly employed. Be-

cause of the possibility of peptization when washing with water it is a

good idea to replace the beaker receiving the filtrate with another emptybeaker just before the final stages of washing. Then, if leakage does occur,

the volume to be refiltcred will not be so large. (See p. 240.)

The washing of a precipitate must be continued until a test on a few

milliliters of the filtrate which has just filtered through reveals no presenceof contaminating ions. During washing it is advantageous to fill the filter

about three-quarters full of wash solution and allow this to drain practi-

cally completely before refilling to the three-quarters level. Rapid cleansing

is also promoted if the wash solution is played into the filter so that the

Page 30: quimica inorganica cuantitativa


whole of the precipitate is agitated; care should be taken, however, to

avoid spattering.

Subdivision of Wash Liquid. It has been mentioned already that in

rinsing following the transfer of liquids from one vessel to another, and in

washing precipitates, it is better to use many small portions of the liquid

than to use a few large portions. After the addition of wash solution the

quantity of contaminant in the solution still clinging to the precipitate,

for example, may be calculated as follows:

Let W be the milliliters of contaminating solution adhering to the

precipitate and let L be the milliliters of wash solution available for each

washing operation. Let Co be the concentration of the contaminant in

g./ml. in the original W milliliters, and Ci, C 2 . . . C n be the concentra-

tions of the substance adhering to the precipitate after the first, second

. . . nth washings.

After the first washing the concentration of the contaminating sub-

stance in the liquid adhering to the precipitate is obviously

(i) CI =W~TL' CO

and following the second washing with another L ml.

(2) C 2- w-^ ,


Substituting the value of Ci from equation ( 1 ) into equation (2),

r - W W _[ W_ I2

L" ~W~+L

' W+T ''~ ~


nncl, similarly, after n washings

0) ^

For example, suppose that 1 ml. of the supernatant liquid adheres to

the precipitate after filtering and that the concentration of the objection-

able ion is I g./ml. Assume that the washing were done with one portion of

wash liquid of 50 ml. volume. In this case the concentration of the solu-

tion clinging to the precipitate would be

whereas if the wash liquid consisted of the same total volume of 50 ml.

but was divided into five portions of 10 ml. each, and these were succes-

sively used,

Page 31: quimica inorganica cuantitativa


The increase in efficiency is therefore


The rule of many small portions of wash liquid in preference to few large por-

tions is similar to the principle involved in the process of extraction. Sometimes it

is expedient to recover a certain solute from an aqueous solution, for example, by

shaking the water solution with an immiscible (usually organic) solvent in which

the solute is much more soluble than in water. The ratio of the solubilities of the

solute in the two solvents is the theoretical distribution constant. To illustrate, a

substance might dissolve to only half the extent in one solvent as in another; the

distribution constant would have a value of 0.5 and if equal volumes of the two

solvents were shaken with a given weight (not exceeding the total solubility) of

the solute there would be, after the two liquid layers separated, one-third of the

solute in the one liquid and two-thirds in the other. (K = K * ?3 = 0.5.)

The mathematical proof of the advantage of using many small portions of the

extracting liquid is developed as follows:

Let W ml. of a solution containing X g. of solute be treated repeatedly with

L ml. of extractant. Following the first extraction let Xi be the grams of solute

remaining unextracted. If K is the distribution constant (defined as the ratio of

the solubility in the extracted phase divided by the solubility in the extracting

phase), it may be evaluated thus 4:

U\ K Xl/W(4) K =

(X. - X,)/L

and solving for Xi,

V - 7 W <X - *0 - Y(5) Xi-fvW-

L- A

After the second extraction

KVi(6) X 2

- A,

Substituting in (6) the value for Xi found in (5),

m x x *w KW - \ rAW r

(7) X 2- X r



[KW + LJ

or, in general,"


From equation (8) we may calculate the weight of solute remaining unextracted

after any number of operations. Suppose a liter of aqueous solution containing

10 g. of a solute were extracted with a liter of an immiscible solvent, and assume

that the value of K is 0.5. If the whole liter of extractant were used at once the

3 Assuming that thorough agitation has been established in each operation and that

there are no complicating factors, e.g., adsorption.4Taylor, Treatise on Physical Chemistry, D. Van Nostrand Co., New York, 1931,

pp. 485-87.

Page 32: quimica inorganica cuantitativa


value of Xi would be


On the other hand, if two portions of extractant of 500 ml. each are successively

used, X 2= 2.5 g. Subdivision into ten 100 ml. portions yields Xi = 1.6 g. It is

evident then that the use of many small portions is of much advantage. WhenK differs widely from unity, as usually is true, the advantage is much more

marked than in the example cited.

Equation (8) is often employed to calculate the efficiency of extraction when

L ml. of extracting liquid is to be repeatedly used. In such a case the total volume

of extractant would approach infinity as n became infinitely great; at the same

time X n would approach zero as a limit. This situation, while theoretically inter-

esting, should be distinguished clearly from the practical situation which alwaysdeals with a finite quantity of extractant. This limited amount of extractant, as

the numerical example already given shows, should be subdivided into many small

portions of L/n ml. each. When the total quantity of extracting liquid, L, is

limited, so also will be X, the amount of solute remaining unextracted. We mayemphasize the distinction if instead of equation (8) we write


KVT+L/n \


It has been shown 5 that here the limit of X, as n approaches infinity, is not zero

but proves to be e~L/A'w

. The point has sometimes been overlooked in extraction

work that it is impossible, when a definite quantity of extractant is used, to

approach 100 per cent efficiency ; that, no matter how great the subdivision of the

extractant may be, a portion, namely, e~L/A'w

, of the solute originally present in the

W ml. of solution cannot be removed. This has been experimentally demonstrated

in a number of extraction studies by von Saaf. 6 In brief, equations (8) and (9)

show, when we solve for the limiting values, that repeated extraction with L ml.

portions would lead to the use of an infinite total quantity of extractant, and in

this case the amount of unextracted solute approaches zero as a limit; but when a

finite total quantity of extractant, L ml., is subdivided even into an infinite

number of infinitesimally small portions, the amount of unextracted solute

approaches a finite value as a limit, namely, e~L/K .

Igniting Precipitates. After a precipitate has been filtered and

washed clean it must be dried before weighing. When a Gooch or other

filter crucible has been employed it is customary to put the crucible in a

beaker and then place in a constant temperature oven at a suitable tem-

perature, usually around 105 to 120. The crucible may be placed in a

muffle furnace, provided the temperature is not so high as to cause an

indefinite change in the composition of the precipitate. If a filter paper has

been used it is the usual practice to ignite the paper and precipitate before

the weighing. Certain cautions must be observed. Although many analysts

transfer the wet filter paper with the precipitate directly, to the crucible

for ignition, it is safer first to dry the filter paper somewhat by placing

Griffin, Ind. Eng. Chern., Anal. Ed., 6, 40 (1934).

Griffin and von Saaf, Ind. Eng. Chem., Anal. Ed., 8, 358 (1936).

Page 33: quimica inorganica cuantitativa


the funnel vertically in the oven at 105 for 10 or 15 minutes. The paperand precipitate may then be removedwith less danger of losing any precipitate

than when wringing wet. After the filter

is removed from the funnel it is carefully

pressed flat, the corners turned in and the

top folded over so that the precipitate is

completely enveloped. The bundle is then

transferred to the crucible, which has pre-

viously been weighed to a constant weight,and the ignition is begun. A triangle of

clay or platinum holds the crucible. Thelid is placed on the crucible and heat is

applied from a small flame of the burner.

After a few minutes the contents of the

crucible are quite dry and the paper should

soon begin to char; if it does not, a slightly

higher flame should be used. As a rule the

charring should require about 15 minutes.

Should the paper begin to flame the burner

must be removed at once so that the flame

is extinguished. When no more smoke is

observed the crucible is tilted and the

FIG. 4. Ignition of filter paperand precipitate.

cover partly removed as seen in Fig. 4. The flame is now elevated but

only the blue oxidizing portion of the flame should

touch the crucible. After about half an hour all of

the carbon should have disappeared and the crucible

and precipitate are ready to be cooled.

Use of the Desiccator. After the ignition the

crucible is transferred with tongs to a desiccator to

cool. If allowed to cool in air, considerable moisture

would be taken up. The desiccator (Fig. 5) providesan atmosphere which should be of constant and low

humidity. After not less than 20 minutes, nor muchmore than an hour, the crucible is removed from the

desiccator and weighed. It then is heated again for

the same period as the first ignition and again cooled

and weighed. The ignition, cooling and weighing mustbe repeated until two successive weighings agreewithin 0.3 mg. Then and then only may the crucible

FIG. 5. Desic-

cator containing

weighing bottle andcrucible. (Courtesy,

Popoff : Quantitative

Analysis, Philadel-

phia, The Blakiston


and contents be said to be dry and of constant weight.

It should be realized that no desiccant is a perfect drying agent, althoughsome, like phosphorus pentoxide and magnesium perchlorate, are nearly so. The

Page 34: quimica inorganica cuantitativa


most commonly employed desiccants are anhydrous salts which, when they func-

tion as drying agents, take up moisture to form hydrates. Exactly what happenswhen a moist object is placed within a desiccator containing an anhydrous salt

as a desiccant may be understood most clearly from the standpoint of the phaserule7 which states that

P 4. C + 2

where P is the number of phases (gas, liquid, solid) present in a system at equi-

librium; C is the number of components (the smallest number of individually

variable; constituents mathematically defining the composition of each phase); and

F is the number of conditions, usually only temperature, pressure and composi-

tion, which may be varied without causing a change in the number of phases.


AB.SHgpa vapor

TemperatureSchematic drawing of vapor pressure curves for a hydrated salt.

Suppose that we have a salt, AB, which may form the three different hydrates,AB.H 20, AB.2H 2 and AB.3H 2O. What relation exists between the salt and the

moisture in the atmosphere of the desiccator? The anhydrous salt may exist in

contact with water vapor within certain limits of temperature and pressure with-

out the formation of any hydrate whatsoever, even the lowest hydrate. The ac-

companying drawing shows the pressure-temperature relations which must hold

under varying conditions. It shows an area for the bivariant equilibrium, AB +vapor; here both pressure and temperature may be varied simultaneously, within

limits, without a change in the number of phases. That is, F = C + 2 P = 2

+ 2 2 = 2, for we have two components, salt and water, and the two phases,solid and gas. If the vapor pressure is sufficiently increased (as when a moist ob-

ject is placed in the desiccator) a point is finally reached at any given temperaturewhere the lowest hydrate, AB.H 20, begins to form. When this happens there are

three phases present: AB, AB.H 2 and vapor. Therefore F = 2 + 2 3 = 1 andthe system is univariant; the pressures and temperatures for this equilibrium are

represented by curve OD which is called the dissociation curve for the mono-

7 For the derivation of the phase rule see any text on physical chemistry.

Page 35: quimica inorganica cuantitativa


hydrate, AB.II 20. Similarly there is a dissociation curve for each of the other two

hydrates, curves OE and OF. The areas between these curves all represent re-

gions where bivariance prevails; only along the curves themselves do we haveunivariance.

To return to a consideration of dehydrating efficiency: it is obvious that if the

desiccator contains anhydrous AB at a given temperature, 1, and a moist objectis placed in the atmosphere above the desiccant, nothing will happen provided the

moisture present is not enough to create at least pressure 2, and the desiccant will

take up no vapor at all. If suilicient moisture is present initially to cause a pres-sure as great as or greater than point 4, the extent to which the pressure is reduced

will depend upon the ratio of the quantity of salt to the quantity of vapor. If

the amount of salt be just enough so that all of it is converted to AB.3H 2 the

pressure established will be that of point 4; if it is so much that all of the vaportaken up goes into the formation of AB.H 2O and no higher hydrate, then the pres-sure will fall to 2. Under no condition can the pressure fall to zero, save only if the

temperature were absolute zero.

Obviously when a dry object (for example, a crucible which has been heated

to a high temperature) is placed in the desiccator we are not surrounding the

crucible with a "dry" atmosphere. The desiccant immediately after being placedin the desiccator consists of anhydrous salt plus at least some of the lowest hy-

drate, AB + AB.H 2O, and for these two to coexist along with vapor demands, at

a given temperature, no degree of freedom. This means that one pressure and one

only can prevail namely, pressure 2, in the diagram. The dry crucible therefore,

as it cools, will begin to take up moisture and actually gain weight. Because of

this, objects should never be allowed to remain in a desiccator indefinitely. Anhour is the maximum Lime and 20 minutes is enough. That constant weights of

objects can be obtained at all is due to the fact that the rate of cooling is fast

and the rate of phase equilibrium is relatively slow as a rule. Thus the object is

cool before it has had time to reach phase equilibrium with the atmosphere in the

desiccator (which, in turn, is in equilibrium with the desiccant), that is to say,

before it has taken up all of the moisture which it would, given time. This em-

phasizes why it is important not only to avoid an unduly long period for cooling in

the desiccator but always to allot approximately the same length of time for cool-

ing when weighing to a constant weight.Aside from the above theoretical considerations it is important to observe the

practical precaution of keeping the cover on the desiccator at all times exceptwhen transferring an object to or from it. Even then the cover is removed and re-

placed (is quickly as possible in order to reduce to a minimum the amount of moist

air admitted from the atmosphere.8 Furthermore the cover should always be kept

greased with petrolatum or, preferably, stopcock grease.


The subdivision of analytical chemistry known as volumetric analysis

deals' largely"with th measuremSTTtr^ef^lun^

merits 'commSiify (jflrploy^d in llibtype uf wetfe-are lire gi'adfiated cylin-

der, the volumetric flask, the pipet and the buret. The cylinder is used

only for approximate measurements of volume; the last three are used

8 Booth and Mclntyre, Ind. Encj. Chem., Anal. Ed., 8, 148 (1936); Hillebrand, U.S.

Geol Survey, Bull 422, p. 119 (1910).

Page 36: quimica inorganica cuantitativa


when exact quantities of liquid are desired. Volumetric instruments are

made either to contain or to deliver a certain volume. As a rule volu-

metric flasks belong to the former class while burets and usually pipets

are of the latter type. The question of exactness in measuring volumes will

be explained further in the chapter on calibrations but it may be pointed

out at once that great care must be exercised in properly handling these

pieces of apparatus if correct measurements are to be attained.

Volumetric Flasks. The volumetric flask (Fig. 6) is a somewhat

pear-shaped glass vessel having a long narrow neck. It.may or may not be

fitted with a ground-glass stopper. On the neck will be found an etched

ring which indicates the level to which it must be filled with a liquid in

order to contain the volume marked on the body of the flask at the tem-

perature indicated. When the flask is used to make up a definite volume of

a solution the proper weight of solute must be dissolved in a small quan-

tity of the solvent in a beaker; the solution then is poured down a stirring

rod from the beaker into the flask. The beaker is rinsed several times with

the solvent and the rinse is carefully added to the flask. More solvent

then is added to the flask until almost to the neck, after which the flask

is whirled, care being taken not to wet the neck above the calibration

mark. The solution then is allowed to attain room temperature, which

should be within two or three degrees of that for which the flask is cali-

brated, after which solvent is added until the meniscus is tangent to the

mark. The flask then is stoppered and the solution thoroughly shaken.

Sometimes a volumetric flask bears two marks, the upper one for

delivery and the lower one for content. When used to deliver, flasks

should be emptied by gradually inclining so that a continuous stream

results, and should be held finally in a vertical position for 30 seconds to

allow for drainage. Then the mouth of the flask should be touched to the

receptacle to remove the hanging drop.

Pipets. A pipet (Fig. 7) is a narrow tube with a bulb near the middle.

The bulb has a capacity slightly less than the volume indicated on the

pipet. An etched ring on the tube above the bulb marks the level for the

liquid in order that the pipet shall deliver (transfer pipets) the volume

specified. (Mohr pipets do not have the bulb but consist of a straight tube

usually graduated in tenths of milliliters.)

A pipet is used in the following manner: After cleaning the pipet


it is rinsed at least three times with the solution to be

9 To test for cleanliness fill the instrument with distilled water and let drain into a

beaker. Examine carefully. If any droplets cling to the inside surface the instrument is

not clean. Cleaning solution is prepared by dissolving 30 g. of sodium dichromate in a

liter of concentrated sulfuric acid. The solution may be used slightly warm whencleaning volumetric apparatus, but never hot, since this causes the glass to expand andunless an equal contraction takes place upon cooling the capacity of the instrument

will be altered.

Page 37: quimica inorganica cuantitativa



/ \

\ x


250ml.20 C




45 Sec.



FIG. 6. (Left). Volumetric flask.

FIG. 7. (Right). Pipet.

measured. This is done by drawing small portions into the pipet through

suction applied by the mouth to the upper end of the pipet. The pipet is

held in a horizontal position and rotated so that the whole inside surface

is wetted with the solution. The rinsings are discarded. The definite vol-

ume of solution now may be measured by filling through suction to a point

above the graduation. Then the forefinger is quickly placed over the top

of the tube so that the liquid is held in the pipet. The pipet is now held

over a spare beaker and a little air is admitted by carefully releasing the

pressure of the finger. When the level of the liquid is tangent to the mark

(viewed in a line horizontal with the eye), the hanging drop is gently

tapped off by bringing in contact with the spare beaker. The pipet then is

held vertically over the receiving vessel and the contents allowed to

drain. About 15 seconds after the flow ceases the tip should be touched

to the surface of the vessel. The small quantity of liquid remaining in

the tip should not be blown out.

Page 38: quimica inorganica cuantitativa


Burets. "Burets (Fig. 8) are used in volumetric analysis in the opera-

tion known as titration. In a titration the solution is run from the buret

into a ves.se! where a chemical reaction takes place. The flow from the

buret is 9 topped when chemically equivalent quantities of reactants have

been brought together. Thus the stoppage may occur at any volume and

burets accordingly are usually marked in tenths of milliliters. A buret

should be cleaned thoroughly10 and rinsed three or four times with the





FIG. 8. Burets.

10 See footnote 9, p. 18.

Page 39: quimica inorganica cuantitativa


solution with which it is to be filled in the same manner as was described

for the pipet. It is then filled, and by withdrawing solution into a spare

beaker the meniscus is allowed slowly to approach the zero mark. There

must be no air bubble in the tip and any hanging drop must be removed.

When the buret is used it should be clamped in a vertical position and

when the flow of liquid has ceased the hanging drop must be removed by

touching the tip with the inside surface of the receiving vessel. The

volume reading should not be taken until 30 seconds after the outflow

has stopped in order to allow for drainage.

= 9

FICJ. 9. Parallax. Only when the eye is on a

level with the meniscus will the reading give no

parallactic error.

Parallax. When using volumetric apparatus the lowest point of the

meniscus should be taken for the reading except in cases where the

meniscus is not observable, as, for example, with permanganate solution.

In such a case the upper periphery is taken for the reading. In any reading

it is imperative to avoid errors of parallax by viewing the instrument

with the eye on a level with the meniscus. If the etched line extends more

than halfway around the instrument the eye is on a level with the gradua-

tion mark if the front and back of the mark arc coincident. It is evident

from Fig. 9 that if the eye is too high the reading will be too high, and vice


Questions and Problems

1 . Name several examples of instances when the sample taken for analysis should

not be representative of the whole body of material.

2. What objections are there to excessive grinding when reducing the particle

size of a sample of material for analysis?

3. Discuss the steps necessary to obtain a sample, suitable in size and in degree

of fineness for analysis, from a carload of coal.

4. Name the fluxes most commonly used for converting acid-insoluble substances

into soluble form.

Page 40: quimica inorganica cuantitativa


5. Explain how the digestion of a precipitate will result in a larger particlesize of the

"insoluble" substance.

6. Why are the"ashless

"filter papers used in quantitative work more expensive

than the filter papers used in qualitative analysis?

7. Explain why a precipitate which has been filtered does not leak through the

paper until it has been washed clean with water, but then may begin to run

through. How may the leakage usually be prevented?

8. Prove mathematically that in washing a precipitate the subdivision of a

given quantity of wash liquid into many small portions which are used suc-

cessively increases the efficiency of the washing.

9. What is the objection to leaving a dried object in a desiccator for a long time

for example, overnight before weighing? Explain fully.

1 0. Which has a greater capacity, a 250 ml. volumetric flask marked "to deliver"

or a 250 ml. volumetric flask marked "to contain"? Explain.

11. What is the best method of testing a buret for cleanliness?

Page 41: quimica inorganica cuantitativa

Chapter 3


J.HE PURPOSE in making any observation or measurement is to learn

the truth about the magnitude involved. Except by accident this maynever be wholly realized, and even if it were we would be ignorant of the

fact. Exactly what distance separates two points, X and Y? One mayapply a measuring stick to ascertain the distance. The experimenter maycheck himself by making the measurement several times. Each time he

will get a somewhat different result. His results will be more concordant

the more careful he is in laying off the lengths of the ruler time after time;

furthermore, they will be better if his eyesight is good than if not. An-

other individual may repeat the measurement and the results may more

or less coincide with those of the first observer but, except by accident,

will not exactly duplicate them. It is evident also that both observers

will of necessity fail to perform a good job if the measuring stick is not

accurate. It follows therefore that as far as truth is concerned we cannot

surely know the distance between points X and Y. What we can do is to

procure the best measuring stick possible, check that stick against some

standard linear measure which is thought to be highly dependable, work

at a constant temperature, have several persons carefully make the

measurement, eliminate any arithmetical errors of computation (here

only addition) and finally take an average of these several determina-

tions. In all probability then this average or mean should be close to the

truth regarding the distance between X and Y. But actually no one ever

can say that this or that result, or that the mean of several measurements,

is the truth, i.e., that the error is zero. The most of which we can be cer-

tain is that we have in our mean an approximation which, because of all

the precautions taken, closely approaches the truth.

It must be realized at once that because two or more determinations of

the same quantity check one another closely we have no guarantee of

small error. If the meter stick with which a certain distance is measured

is, say, actually 100.1 cm. in length, erroneous results inevitably must be

obtained regardless of the fact that duplicate measurements may agree

closely with one another. This points up the distinction between the terms

precision and accuracy.23

Page 42: quimica inorganica cuantitativa


Precision and Accuracy. Precision may be defined as the narrowness

of the limits between which it is thought the true value lies. It is a meas-

ure of the agreement between two or more determinations of a given

quantity. On the other hand, accuracy, a term sometimes mistakenlyused interchangeably with precision, denotes the concordance between the

result of a measurement and the true value of that measurement. Strictly

speaking, since the true value is unobtainable with absolute certainty,

this concordance cannot be gaged. Nevertheless, with proper care and

proper instruments the true value of a measurement may be obtained

within such a small possible span of error that one may assert that the

overwhelming probability is that the true value lies within a narrow rangeof which the reported value is the mean.

Obviously if the same error is made to about the same extent in dupli-

cate determinations, the precision would be high while the accuracywould be low. Accurate results are necessarily precise but precise measure-

ments may or may not be accurate. By custom, precision is expressed in

terms of deviation (usually in parts per thousand) from the mean of the

several determinations; accuracy is estimated in terms of error, i.e.,

departure from the true value.

Practically speaking, accuracy may often be attained (the truth be

expressed) if the precision attempted is not too high. To illustrate: Thedistance between X and Y might be measured by a number of able per-sons with a steel meter stick which has been calibrated to the millimeter

by the Bureau of Standards. Precautions such as regarding the tempera-ture of the surroundings and computing the consequent correction for

expansion of steel are made. If an average result of 33.52 m. is reported,the overwhelming probability is that this is the truth; it is accurate. It is

hardly reasonable to think that the true value differs from 33.52 m. bymore than 1 cm. under the circumstances, particularly if all of the several

determinations fell within, say, 0.005 m. of the average value, and if,

furthermore, the greatest estimated error due to all possible sources adds

up to less than 0.005 m.If a somewhat careless determination were made in triplicate by

another person with results of 34.11, 34.12 and 34.32 m., we conclude that

his average of 34.18 m. is somewhat precise but not very accurate. The

precision is computed as follows (note that the sign of the deviation is

ignored) :

Deviations from Mean





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The absolute mean deviation of a single measurement is 0.09 in a total

of 34.18; the relative deviation is

0.09:34.18 = XrlOOOX = 2.6 parts per 1000

The accuracy of the above determination may be said to correspondto an absolute error of 34.18 33.52 = 0.66 m. This would be, in relative

terms, 20 parts per 1000. The precision in this case is over seven times the

accuracy. Precision then measures the rcproducibility of measurementswhile accuracy indicates the approach to the actual correctness or truth.


Though no one may hope to attain absolute perfection in makingmeasurements, it is profitable to learn as much as possible about the typesof errors likely to occur so as to be in a position to reduce their effect to a

minimum. Errors are often divided into two classes: determinate errors

and indeterminate (accidental) errors.

Determinate Errors. These arc the errors which persistently recur

from one determination to another. Their magnitude thus can be esti-

mated fairly well and corrections for them applied. If the error is of the

same magnitude time after time (as, for example, would result from the

repeated use of a weight supposedly 50.0000 g. which actually was 50.0011

g.) it is called a constant error. Often the direction and the degree of this

type of error are known and appropriate corrections may be applied.

Among the common determinate errors the following may be listed:

1. INSTRUMENTAL ERRORS, the result of poor construction of appa-ratus, faulty calibration of weights or volumetric apparatus, impurereagents, etc.

2. PERSONAL ERRORS, the result of idiosyncrasies of the individual,

such as inability to note color changes sharply in persons whose sense of

color is poor. Such errors usually are constant both in degree and direc-

tion. Under this heading comes the not uncommon error of matchingcolors of a given indicator in titratioiis or in colorimetric determinations,even in persons whose eyesight is perfect but who are prejudiced by the

hue of color obtained in a previous determination. If a certain depth of

color has been obtained in the first experiment, it is easy to convince one-

self, if the buret reveals a reading slightly lower for the duplicate deter-

mination, that one or two more drops of reagent are needed in order to

match exactly the color first obtained. Such prejudices can be eliminated

by deciding definitely about the color before taking the reading.3. METHODIC ERRORS, which come about for various reasons, all

having to do with problems inherent in the method of analysis, from the

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sampling (which theoretically can seldom be perfect) on through to the

final operation. Practically all analytical methods have their own charac-

teristic inherent errors. These are the most serious of all errors for theycan be eliminated or minimized only through working out a different

procedure. Other methodic errors may be due to failure of a reaction to

take place quantitatively, coprecipitation, slight solubility of precipitates

in gravimetric determinations, inability to dry a precipitate completelyand the like.

At least some methodic errors may be minimized by better technique.

For example, there are usually means of diminishing coprecipitation.

Errors due to solubility will be less if, in washing precipitates, one uses

the proper wash solution and in the smallest, yet sufficient, quantity. In

any event careful attention to the possible type and probable magnitudeof these errors is the responsibility of the analyst.

Indeterminate Errors. These errors are"accidental" in the sense

that they occur for no apparent cause even when the determination is

carried out by a competent analyst and under conditions which are

thought to be invariable. The perspicacity of the analyst is of no avail in

the case of true, indeterminate errors, and the theory of probability is

used to cope with them. In spite of the fact that corrections to counteract

the effect of accidental errors cannot be applied it is possible to arrive at

an intelligent conclusion regarding what is the "best" number to repre-

sent the result in a series of measurements. If an infinite number of ob-

servations were made the frequency of the occurrence and the magnitudeof the accidental errors may be expressed by the probability law. The

frequency of the occurrence Y, of an error of magnitude X, is given by the


(1) Y = -=VTT

where Y = frequency of the occurrence of the deviation

X = magnitude of a deviation of frequency of Yh = a constant the value of which depends upon the charac-

ter of the measurements (thus its value is the same onlyfor a given kind of measurement)

e and TT = constants, 2.718 . . . and 3.141 . . . , respectively.

The curve represented by this equation is called the probability curve

or the error curve and is shown in Fig. 10. Three conclusions are readilydrawn from the qualitative nature of this curve: small errors occur moreoften than large ones; very large errors are very unlikely; and positive

Page 45: quimica inorganica cuantitativa


errors and negative errors of the same magnitude are equally likely. Thusin a series of reliable measurements it follows that the value most closely

approaching the true value will be one such that the sum of the differences

between it and the larger individual measurements will be equal to the

sum of the differences between it and the smaller individual measure-

ments. Such a value is called the arithmetic mean.

FIG. 10. The probability (error) curve. (Courtesy,

Popoff: Quantitative Analysis, Philadelphia, The Blakis-

ton Company.)

Average Deviation of a Single Measurement. The validity of a

given single measurement may be expressed in terms of the magnitudeof the deviation that probably attaches to it. This is simply the averageof the several differences between the individual measurements and their

arithmetic mean, or

(2) Av d _ Ld 'I+


XV V 14.'


where di, d . . . dn are the individual deviations from the mean.

Average Deviation of the Arithmetic Mean. This term should be

distinguished from the above. The average deviation of a single measure-

ment is suitable to give us an estimate of precision as was done in a pre-

vious illustration (p. 25). It tells us the average extent by which the

several individual determinations differ from their mean. But to what

probable extent is this mean itself subject to error; that is, how greatly

does it probably depart from the truth?

Before answering this question let us realize that if we take the meanof a few measurements we have a less reliable approximation than if we

take the mean of many measurements. It may be shown by the Method

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of Least Squares that the reliability of the arithmetic mean of a series

of n measurements is \/n times as reliable as a single measurement.

For example, the mean of 16 measurements is 4 times as reliable as a

single measurement and that of 4 measurements is twice as reliable as

one measurement. If we designate the error of a single measurement by

ii, then the error of the arithmetic mean of n measurements is u/\/n,

and the error involved in the arithmetic mean of 4 individual measure-

ments is obviously twice as great as that involved in the mean of 16


To come back to the earlier question, we may say then_thatthe

arithmetic mean is probably in error to an extent which is l/\/n as great

as the average deviation of a single measurement, or

(3) Av. Dev. of Arith. Mean = '~Vn

Thus if 16 measurements gave a mean of 0.10126 and an average devi-

ation for a single measurement of 0.8 part per 1000, then the average

deviation of the arithmetic mean would be 0.8/\/16 = 0.2 part per 1000.

This means that the value of 0.10126 is possibly in accidental error by0.00002 or that the true value should be not less than 0.10124 nor greater

than 0.1 0128.

Standard Error. There are still other ways of expressing errors. The

mean error of a single measurement, often called the standard deviation of a

single measurement, is defined 1 as the error the square of which is the

mean of the squares of the individual deviation from the arithmetic

mean, without regard to sign, or

/d~2Stand. Dev. of Single Measurement =


The above, of course, is rigorously valid as a limiting value for an infinite

number of observations; if the number of measurements is small a closer

approximation of the standard deviation may be calculated by using

n 1 instead of n in the denominator under the radical, i.e.,

n - 1

Probable Error. The probable error of a single measurement is that

error of such magnitude that in an infinite number of measurements

the number having larger errors than it will be equal to the number

having smaller errors than it. The equation for the probable error is

l But see Razim, ./. Chem. Education, 19, 411 (1942).

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(5) Probable Error of a Single Measurement



n - 1

When ft = c the relation between the three measures of unrelia-

bility are:

Av. Dev. = 0.8 Mean (Standard) Error = 1.18 Probable Error

It will be recalled that the expression giving the average deviation

of a single measurement, namely, equation (2), if divided by \/n9 yields

the average deviation of the mean. Similarly, by dividing equations (4)

and (5) by \/n we obtain, respectively,

Stand. Dev. of Arith. Mean fii*_+. d 22 + dn



rW(Mean Error of Arith. Mean)

~\ n


/ V "

(7) Probable Error of Arith. Mean = 0.67 J^-^^~I~~**

"/^The application of these six forms may be illustrated by the follow-

ing example. It should be remembered, however, that any equationbased upon the Probability Law presupposes a large number of measure-

ments, and that with a small number these equations do not commandthe rigor attained if the calculations could be based upon an unlimited

number of observations.

Problem. The results of three measurements were: 55.09, 55.14 and 55.15.

Calculate (A) the average deviation, (B) the standard deviation and (C) the

probable error of a single determination, and (A7

, B 7 and C') of the mean.


Results Individual Values of d d 2

55.09 0.04 0.0016

55.14 0.01 0.0001

55.15 0.02 0.0004

Mean 55J3 2d 007 id 2 6.0021

(A) Av. dev. of single determination: - *--- = 0.023 absolute, orj

0.023:55.13 = X:1000; X = 0.42 part per 1000.

023(A') Av. dev. of arith. mean: -^-= = 0.013 absolute, or 0.24 part per 1000.

V3(B) Stand, dev. of single determination: /0.0021

(Mean error of single determination) \~~T~" absolute, or 0.58

part per 1000,

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(B') Stand, dev. of arith. mean: 0.032 = Q Q18 absolute>or 35 part per 1000 .

(Mean error of arith. mean) -y/3

(C) Probable error of single determination: (0.67)(0.032)- 0.021 absolute,

or 0.38 part per 1000.

(C') Probable error of arith. mean: -^ - 0.012 absolute, or 0.22 part per


Number of Measurements. The question of how many times an

analysis should be carried out in order to have full confidence in its

result varies with the nature of the particular analysis.. It is customary

to run a primary standardization of a solution at least in triplicate; the

analysis of an unknown substance usually is done in duplicate or in

triplicate depending upon the certainty with which the composition

must be known. In very precise work such as an atomic weight determi-

nation as many as 10 parallel determinations may be made. Recalling

that the accidental error of the mean in a series of measurements varies

inversely with the square root of the number of measurements, and

bearing in mind that the error usually is small in comparison with the

magnitude of the mean itself, we realize that there must come a time

when the small, added reliability resulting from still further determi-

nations will not repay the added effort. Let E be the accidental error of

the arithmetic mean for n determinations. Then the reliability of the

result expressed in terms of the accidental error will vary as follows:

No. of Accidental Error of

Determinations Arithmetic Mean'

n E2n E/V2 = 0.71E

3/i E/V3 = 0.58E

4n E/V4 = 0.50E

5n E/\/5 = 0.45E

10/i E/VlO = 0.32E

lOOn E/vlOO = 0.10E

Obviously it would be a rare occasion indeed when it would be

practical to increase the number of determinations a hundredfold if the

reliability would be increased only tenfold.

Rejection of a Result. When several parallel determinations have

been made, sometimes one result is found to differ considerably from

the others. If the analyst is unaware of any reason for the discrepancy,

the question arises as to whether or not the suspected result legitimately

may be eliminated before computing the mean result. No categorical

answer can be given to meet all such cases. However, it can be asserted

that there must be a critical value which the suspected result cannot

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exceed without the probability of its being a valid member of the series

decreasing to the vanishing point. Reliable results will not deviate from

the arithmetic mean beyond certain limits. We have seen that large

accidental errors are very unlikely and it is extremely improbable that

many small inherent errors should all assume the same sign in a givendetermination. More likely a doubtful value results from a determinate

error though the analyst may be unaware of any mistake having been

made. A low result in the gravimetric determination of chloride mightbe due to an unobserved loss of precipitate, for example.

A good many rules have been stated regarding the extent to which a

suspected result must deviate from the mean in order to warrant re-

jection. One rule often followed in quantitative analysis states that if a

total of four or more determinations have been made, including the sus-

pected one, the latter may be omitted in computing the mean providedits deviation from the mean of all the others exceeds four times their

average deviation. While it may be proved from the theory of probability

that one is justified 99.3 per cent of the time if this rule is followed, the

rule often allows the inclusion of a result which is likely to be in such

error that its inclusion is fully as objectionable as the unjustified ex-

clusion of a result. Therefore, the following rule laid down by Kolthoff

and Sandell2 and based upon the criterion of a number of statisticians

seems preferable, namely: Discard a result if its deviation from the arith-

metic mean of the other results exceeds twice the standard deviation of a

single measurement. This rule is valid to an extent of 95.5 per cent; i.e.,

the odds are 95.5 to 4.5 or 21 to 1 that a result discarded according to

this rule is of a determinate nature and not a matter of chance. Therule should be employed only if the number of determinations is at

least 4, but not much greater than 10, certainly not greater than 22;

in a series of 22 determinations one result will normally show a devia-

tion as great as or greater than twice the standard deviation. Actually

in quantitative analysis one usually does not have more than four or

five results from which to strike a mean.

It will be recalled (p. 29) that the ratio of the average deviation to

the standard deviation of a single measurement is 0.8:1, which is to

say the standard deviation is 1/0.8 times the average deviation, or 1.25

times the average deviation. Therefore, the practical application of the

above rule to a given doubtful result is that the result is rejected if

Deviation of doubtful result 0/1 -, c-7 , . ?i ; i > <6(1.4O) = 4.5Average deviation of a single measurement

and we may thus employ the more quickly obtained average deviation

2 Kolthoff and Sandell, Textbook of Quantitative Inorganic Analysis, The Macmillan

Co., New York, 1943, p. 276.

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instead of the standard deviation in drawing a conclusion regarding the

possible rejection of a result.

To illustrate: suppose in an analysis the results obtained in percent-

ages are: 60.64, 60.50, 60.47 and 60.56.

Results Deviations

(60.64), doubtful (0.13)



60.50 0.01

60.56 0.05

Mean 60J31 Mean 0.033

Dev. of doubtful result 0.13

Av. dev. of single result 0.033

Therefore reject 60.64

= 3.9 > 2.5

It is worth repeating that no rule, when applied to a finite number of

results, is perfect, since the theoretical considerations which form the

basis for any such rule rest upon the assumption of an infinite number of

measurements. In actual practice therefore we cannot hope to discover

and apply a perfect rule for rejecting suspicious results; nevertheless,

because it is necessary to deal with a finite (and usually a rather small)

number of results, the inclusion of an abnormal result would exercise a

disproportionate influence on the mean. The mean result would then be

less reliable than if the doubtful one had been rejected. The employ-ment of a practical rule such as stated above is permissible and usually

advantageous because of the rather small number of observations prac-

tically possible.

Minimizing and Correcting for Errors. It is no more than ordinary

common sense for the analyst, when high accuracy is wanted, to use

calibrated apparatus and pure reagents and to employ careful techniquein the course of an analysis. The custom of running a determination in

duplicate or better leads, we have seen, to greater reliability than if a

single result is trusted. In addition there are other means of securing

greater dependability in the results of our work. Two of these will be

taken up.RUNNING A BLANK. A blank is run when it is impossible or impracti-

cable to avoid the introduction of an impurity during the course of an

analysis. For example, if the alcohol itself, used as a solvent for the

benzoic acid employed as the standard in the titration of a base, contains

some acid a blank must be run to determine the acidity of the alcohol.

Then a correction for the titration of the base is applied, the magnitudeof the correction being obtained through the blank run. If the titration

Page 51: quimica inorganica cuantitativa


of a given weight of benzoic acid dissolved in alcohol required 30.00

ml. of base, and a blank run revealed that the acid in the alcohol itself

required 0.15 ml. of base, then the benzoic acid obviously was equivalentto only 29.85 ml. of base. In general, a blank run omits the essential

substance (the sample; or, in a standardization, the standard itself);

all other substances and all quantities and conditions are kept the sameas in the analysis proper.

RUNNING A CONTROL. Suppose that there are inherent errors in a

given procedure for an analysis. The results of analysis by such a methodwould always be in error. If, however, a control run is made, the propercorrection may be applied to the result. The control consists of a determi-

nation in which a known amount of the constituent being sought is

analyzed under the same conditions of concentration, temperature,method of analysis, etc., as the unknown. Inasmuch as the result whichshould be obtained is known in advance, the error may be ascertained

by comparison with the actual result of the control analysis. Thereafter,the proper correction on the result obtained for an unknown may bemade. It must be kept in mind that here again the value of a control is

largely dependent upon "other things being equal"; that is, both quali-

tatively and quantitatively, the sole difference between the analysis of

the unknown and the "analysis" of the control substance should be the

use of an unknown on the one hand and that of a known constituent onthe other. In order that the control may duplicate the complete situation

regarding the unknown it may be necessary to synthesize a control

mixture of approximately the same composition that the unknown is

believed to have.


In making any observation one and only one uncertain figure should

be recorded. (In very exact calculations two doubtful figures are carried

until the final result is computed; then the result is rounded off to onedoubtful digit.) For example, in reading a buret calibrated in tenths of

milliliters the reading is made to the hundred ths; in observing the swingsof the pointer of a balance over its scale which is marked off in units the

reading is made to the tenths of a division, etc.

There are good reasons for emphasizing the importance of the properretention of significant figures. The number of figures employed to expressa certain quantity should tell something of the precision involved. Whenmore digits are used than are warranted the implication becomes auda-

cious, since the user gives the impression of exceeding the limitations of

refinement of the apparatus from which the observations were taken.

When a lesser number of digits than are warranted is employed the

analyst is denying himself of his rights; this indicates carelessness on his

part or ignorance of his rights.

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A number is an expression of quantity. A figure or digit is any of the

characters: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A significant figure is one which

indicates the magnitude of the quantity in the place (i.e., the units

column, the tenths column, etc.) in which it stands. The number 123

indicates that there are one hundred, two tens and three units, with some

doubt as to the exactness of the three units. All of the digits are significant.

The zero is employed in two ways, either as a significant figure or only to

locate the decimal point. If the magnitude of the quantity in the place

where the zero is located is nearer zero than any finite value, then the

zero is significant. Thus the zero is significant in the number 1023, for

this number states that there are one thousand, two tens, and three

units, but no hundreds. Any and all zeros located between two finite

digits must always be significant zeros. In the number 0.00123 none of

the zeros is significant since they merely locate the decimal point. Thelocation of the decimal point in any measurement is governed by the unit

in which the measurement is expressed. Zeros which are not significant

may be eliminated, therefore, simply by using a smaller or a larger unit.

They also may be avoided by using exponential numbers. Instead of the

number 0.00123 meter, the number 1.23 mm. or 1.23 X 10~ 3 m. maybe employed. Instead of the number 15,500 mm., either the number 15.5

m. or 15.5 X 10 3 mm. may be used if, indeed, the zeros are not significant.

The quantity 15,500 mm. illustrates the cases in which there is no waywhatsoever to tell whether or not the zeros are significant. It may be

true that the distance between two points is exactly 15,500 mm. in

which case the two zeros are significant. On the other hand it may be that

the measurement was made with a precision only of 1 m.; in that case

neither of the zeros is significant. If the measurement actually was madewith a precision of 1 mm. the quantity should indicate this by beingwritten either 15,500 mm. 1 mm., or 15.500 m. Both of these numbers

correctly denote that the distance was measured to within a thousandth

of a meter (i.e., to within 1 mm.) in accordance with the rule that one and

only one uncertain digit here the second zero is recorded. In other

words the measurement is certainly nearer 15.500 m. than it is to either

14.499 or 15.501 m. To circumvent all ambiguity it is usually best to

choose the proper unit, that is, a unit which will automatically eliminate

zeros which are not significant.

Computations. The following rules should be observed for recordingdata and for making calculations from these data.

1. As many digits should be used in any observation (i.e., any physical

measurement) as will give only one uncertain or doubtful figure. In

lengthy and very exact calculations it is desirable to retain, until the final

result, two doubtful figures.

2. In rejecting superfluous figures (i.e., in rounding off) increase by 1

Page 53: quimica inorganica cuantitativa


the last figure retained if the following rejected figure is 5 or more. If

the rejected figure is less than 5 the last figure retained is not altered. It

is not necessary to consider at all any digits to the right of that one which

is being dropped. For example, 1.235 and 1.2349 rounded off to the

hundredths become 1.24 and 1.23, and in the case of 1.2349 the figure 9

need not be considered at all.

3. In addition and subtraction all numbers should be rounded off to

an extent corresponding to the first doubtful digit to the right of the decimal

point in any of the terms. The sum of the terms: 0.0211, 52.18 and 6.05891,

assuming the last digit in each case to be uncertain, is





It is useless to extend the digits in any term further than the hundredths

column since the second term has its doubtful figure in that column. Thefirst and third terms are made 0.02 and 6.06 to conform to Rule 2. The

validity of Rule 3 is obvious if we assume that, had we been able to secure

significant figures to the hundred-thousandth place for all three terms,

they necessarily would have been respectively between the values given

below, for otherwise they would never have been read as 0.0211, 52.18 and6.05891 as actually they were.

Smallest possible Largest possible

1st term 0.02105 0.02114

2nd term 52.17500 52.18499

3rd term 6.05891 6.05891

58.25496 Rounded off = 58.25 58.26504 Rounded off = 58.27

Therefore, if rounded off in the first place, before adding, we do not offend

our precision since the value 58.26 does not differ by more than 1 in the

hundredths column as compared with the minimum and maximum of

58.25 and 58.27.

4. In multiplication or division, each factor and the product or

quotient may have no greater precision than the least precise term. It

must be stressed that here it is the relative and not the absolute uncer-

tainty that is transmitted to the result. It therefore is not always correct

simply to round off each factor to the same total number of significant

figures as contained in the least precise factor, though often such a rule

happens to work. It is safer to proceed as follows: Note the factor havingthe smallest total number of significant figures and round off all others to

the same number of significant figures, unless in so doing a number is

Page 54: quimica inorganica cuantitativa


obtained which, ignoring the decimal point, is numerically smaller than

the least precise factor; in that case, retain one more significant figure.

Suppose we wish the product of (7.765) (5.2) (1.21). The number 5.2 has

only two significant figures so that it is the least precise factor. Its true

value might actually be 5.1 or 5.3. This number then could be in error by

0.1 in 5.2 or roughly 20 parts per 1000. Therefore, neither of the other fac-

tors nor the final product may exceed this precision. It is correct to round

off the number 7.765 to 7.8, the same number of digits (two) as the govern-

ing factor, since 7.8 implies a possible deviation of not more than 0.1 in

7.8 or about 13 parts per 1000, and this is better than the precision of the

least precise factor. If, however, the last factor, 1.21, also is cut back to

two significant figures and written as 1.2, the precision is less than that

of the governing factor, for the number 1.2, with a possible uncertainty

of 0.1, might have a deviation of 0.1 part in 1.2 or about 83 parts per

1000. However, if we keep three digits in this factor, using 1.21, its pos-

sible deviation is only 0.01 in 1.21 or 8 parts per 1000. Thus the factors

and the product should be rounded off as follows:

(7.8) (5.2) (1.21)= 49

unless the product is to be used in further calculations, in which case it

would be written 49.1.

5. Logarithms should be used for multiplication or division except

for the simplest operations. Observations usually permit the use of four

figures, for with the analytical balance weighings may be made to the ten-

thousandth gram and burets are read to the hundredth milliliter; this

usually results in a measurement expressed with four significant figures.

Although four-place logarithm tables are sufficiently precise for almost all

chemical calculations it should be remembered that a five-place table will

give an antilogarithiu having four digits without interpolation. The use of

five-place tables therefore is recommended. Logarithms automatically

cause the dropping of the superfluous figures which result in long-hand

multiplication. The 10-inch slide rule should not be used in making calcu-

lations since it does not yield a high enough precision ; however, it may be

used as a quick check on computations made with logarithms.

Effect of Errors on Final Result. The reliability of the final result

of an analysis evidently depends upon several factors namely, the errors

inherent in the procedure itself, the care with which the work is done and

the limitations of the equipment and of the mathematical operations

necessary to the calculations. With a proved method of analysis, good

equipment and a competent analyst, the precision attained in manyanalyses should give a deviation of 1 part per 1000 or better. It has alreadybeen shown that in the operations of multiplication and division it is the

relative precision of the least precise factor which governs the precisian of

Page 55: quimica inorganica cuantitativa


the final result. If one factor free from error is multiplied by a second

factor having an error of 5 parts per 1000 the product also must have an

error of 5 parts per 1000. But in subtraction we encounter a situation

which may introduce an error the magnitude of which often is not fully

appreciated. For example, suppose 1007 is to be subtracted from 1027 and

that the last figure in each number is in doubt to the extent of 1. The dif-

ference becomes (1027 1)- (1007 1)

= 20 2. Note that we be-

gin with two numbers each having a deviation of almost exactly 1 part

per 1000, but the difference, 20 2, has a possible deviation of 100 parts

per 1000! True, in subtraction the errors might cancel, but one cannot be

certain of this so that it is necessary to recognize that the result might be

in error to the maximum extent. Evidently the best way to hold this type

of deviation to a reasonable limit is to arrange things so that the quantity

obtained after subtracting is fairly large. If the weight of a precipitate,

obtained by subtracting the weight of the empty crucible from that of the

crucible plus the precipitate, amounts to about 0.4 g., for example, and if

each of the two weighings carries a possible error of 0.2 mg., the weight of

the precipitate itself will be known to within 0.4 rng. This represents a

deviation of 1 part per 1000 in spite of the subtraction involved. This

means that the original sample taken should be of sufficient weight to

yield a precipitate of about 0.4 g., which, in turn, means that in the be-

ginning one should make an estimate of just what is a suitable amount of

sample.The number of significant figures employed in reporting the final result

depends upon two considerations. If the limitations of the procedure or of

the apparatus used are low the result reported cannot imply a high pre-

cision no matter how carefully the analysis is carried out. For example, if

a sample were weighed on a balance reading only to the third decimal

place and amounted to 0.325 0.001 g., the final result of the analysis

necessarily would be subject to a possible error of at least 3 parts per 1000

even though perfect checks had been obtained in duplicate determina-

tions. Not more than three figures could be used in reporting the final

result (only two if the first digit in the final result were, for example, 9).

On the other hand, if two parallel determinations done with precision

apparatus and by a proved method of analysis yield widely divergent

results due to poor technique, the reported result again may not indicate

a high precision. For instance, if the results were 60.52 and 60.31 there

obviously is doubt about the figure in the tenths column, to say nothing

of the hundredths column; thus the mean reported result must be 60.4

and not 60.42. However, the tenths column is not doubtful in two such

numbers as 60.41 and 60.39 although the digits in the tenths column are

themselves different. Simply subtracting the smaller of two duplicate

results from the larger will show the place where doubt begins: 60.41

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60.39 = 00.02. Thus the uncertainty lies wholly in the hundredths

column, and it is proper in this case, judging from the individual results

alone, to report a mean of 60.40 with the four significant figures.

Questions and Problems

1. Two weighings are made on an analytical balance. If the smaller, 0.2652 g., is

subtracted from the larger, 0.2684 g., what is the possible error of the difference

in parts per 1000?

Answer: 63 parts per 1000.

2. In multiplying (8.24) (3.1) (15.29) how should each factor be rounded off and

what is the product?

3. In the analysis of a certain sample the following results in per cent were ob-

tained: 10.64 and 10.50. What mean result should be reported, assuming all

observations could be to four significant figures? Another analyst obtained for

the same substance 10.59 and 10.61. What average should he report?

4. Explain the general conditions of analysis under which a blank should be run;

under which a control run should be made.

5. How many significant figures are there in the following numbers? 1.8 X 10~ 5;

1001.; 0.050; 0.1001; 16000.

Answer: 2; 4; 2; 4; (?).

6. In standardizing a sodium hydroxide solution the weights of benzoic acid

found to be equivalent to a milliliter of the base in four different titrations

were: 0.01224 g., 0.01222 g., 0.01225 g., and 0.01228 g. Calculate the averagedeviation of a single titration. Should any of the results be rejected?

Answer: 0.8 part per 1000; 0.01228 is rejected.

7. Five determinations of iron in a sample of ore yielded the values 30.27, 30.29,

30.32, 30.36 and 30.37 per cent iron. Calculate the average deviation, the

standard deviation and the probable error of a single determination and of the

mean. First decide whether any result should be rejected.

Answer: Reject none; in order named: 1.1, 1.5, 1.0, 0.5, 0.7 and 0.4 part

per 1000.

8. To what decimal place must the samples be weighed in order to have a pre-cision at least as good as 2 parts per 1000 if their weights are respectively about

0.2 g., 2 g., 40 g. and 500 g.?

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JVL_ASS is the quantity of matter possessed by a body. The weight of a

body is the measure of the force of gravity upon the body. Since the force

of gravity varies with different localities the weight of a body also varies

from place to place; the mass, however, is independent of locality. In

spite of the distinction the term weight is often used to imply mass; this

is so common that special mention usually must be made of the point if it

is desired to emphasize mass in contradistinction to weight.


The analytical balance is the instrument used to weigh a body. The

balance, more than any other piece of apparatus, is indispensable to ac-

curate work. It is imperative, therefore, that we know the balance:

recognize the essential parts and their functions, appreciate the delicacy

of the mechanism and understand how to operate properly this valuable

laboratory tool.

Fig. 11 gives a diagram of the ordinary analytical balance. The vari-

ous parts should quickly come to be recognized and known by their cor-

rect names. There is no substitute for an actual demonstration to becomefamiliar with the working parts of the balance. The laboratory instructor

may partially dismount the balance in order to show its parts and to ex-

plain the necessity of treating the instrument with the respect due anysuch delicate equipment.

The balance is encased in either wood or metal and has glass windowsat the front, back and sides, and usually at the top. The case is mounted

upon either three or four legs of which at least two are adjustable in

length so that the balance may be properly leveled. A spirit level or some-

times a plumb bob is found inside the case. The base of the balance is of

black plate glass and upon it is mounted a heavy column, usually of

brass, which supports the beam. A piece of plane agate which is inlaid in

the top of the column presents a rust-proof surface to an agate (steel in

cheaper balances) knife-edge attached to the center of the beam; it is this

contact between the two agates upon which the beam oscillates when the

balance is in use.

There are also similar agate-agate contacts between the two stirrups

and the extremities of the beam. Thus, as in a fine watch, friction is re-


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FIG. 11. Essential parts of the analytical balance. (Courtesy, Popoff: Quantitative

Analysis, Philadelphia, The Blakiston Company.)

duced to a minimum by these"jeweled" contacts. The beam itself is

graduated either on both sides or only on the right-hand half. The

graduations designate milligrams of \veight, and there may be either 5, 6,

10 or 12 mg. divisions on the arm. A small wire weight called the rider

straddles the beam and indicates according to its location the number of

milligrams of weight taken for a given weighing. The position of the rider

may be changed by lifting it with the rider hook and horizontally sliding

the rider rod.

The balance pans are hung by bow wires from the stirrups. A pointer,

also called the indicator or needle, is attached perpendicularly to the

center of the beam, and when a weighing is being made it oscillates over a

numbered scale in front of the base of the column. A gravity weight,

adjustable in height, is fixed upon the pointer; its height regulates the

period of oscillation of the pointer and the sensitivity of the balance.

A key or knob located front and center to the case, controls, throughthe hollow column, the beam support. This support will, when the knobis turned clockwise, lift the beam about a half millimeter, thus breakingthe three agate-agate contacts and preventing the danger of chipping the

knife-edges if the balance is accidentally jarred while not in use.

Underneath the pans are two pan arrests which push up against the

pans with a gentle pressure. They are controlled by the push-buttonlocated at the front and near the center of the case. The pan arrests keep

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the pans in a plumb position and reduce the likelihood of disturbing the

alignment of the beam when anything is being placed upon the pans.


The foregoing discussion deals with the ordinary analytical balance.

During the past quarter century various changes have been made in

design which facilitate more rapid or more convenient weighing, and in a

few instances contribute to a more precise weighing. Such balances maybe purchased at somewhat higher cost.

Chainomatic Balances. In this type of balance one end of a small

gold chain is attached to the right-hand half of the beam while the other

end is attached to a vernier. The vernier may be moved up and down,thus varying the weight suspended from the beam. The vernier gives the

reading in grams for the second, third and fourth decimal places and does

away with handling of weights except those of hundred-milligram de-

nominations. In some chainomatic balances the use of all fractional

weights is eliminated. The sole advantage of these balances is the rapidity

of weighing achieved, first, because fractional weights are not handled,

and secondly, because the vernier may be elevated or lowered without

stopping the swinging of the beam. Fig. 12 shows the characteristic fea-

tures of chainomatic balances.

Keyboard Balances. Fig. 13 shows a keyboard type of balance.

Fractional weights up to one gram are suspended behind the right-hand

pan so that they may be attached to or removed from the right stirrup by

FIG. 12. Details of the chainomatic attachment. (Courtesy, Popoff: Quantitative

Analysis, Philadelphia, The Blakiston Company.)

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FIG. 13. Keyboard balance. (Courtesy, Fisher Scientific Co.)

operating keys located at the front of the balance case. In a few instru-

ments these mechanical devices have been extended to handle all of

the weights. Here too the chief advantage is speed and convenience of


Damped Balances. Almost any balance may be equipped with a

device which causes the pointer to come to rest on the first swing rather

than to oscillate back and forth. Such damping is attained by means of

oil or air chambers, or by magnetic means, as shown in Figs. 14 and 15.

In the latter case an aluminum vane is attached to the end of the beamor is suspended from the stirrup in such a manner that, as the beam

swings, the vane moves between the poles of a magnet. Such devices save

time in that the position on the scale over which the pointer comes to

rest is noted immediately.

Weights. In making a weighing we are simply assembling a combina-

tion of masses (weights) equal to that of the object being weighed. Aset of weights therefore is required for each balance, Such a set is pictured

Page 61: quimica inorganica cuantitativa


14. E. H.


FIG. 15. Magnetic-damped balance. (Courtesy,

Christian Becker.)

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in Fig. 16. In almost all student sets the following denominations are

found: 50, 20, 10', 10", 5, 2', 2" and 1 g. weights, and 500, 200, 100', 100",

50, 20, 10', 10" and 5 mg. weights. Smaller weights than the 5 mg. weight

are not necessary; in fact no fractional grain weight should be used which

is smaller than the greatest number of milligrams which may be had from

the rider.

Hardly any given weight is perfect. A weight which, for example, the

manufacturer has marked as a 10 g. weight, might in truth be a milli-

FIG. 16. A set of analytical weights. (Courtesy,

Popoff: Quantitative Analysis, Philadelphia, The Blakis-

ton Company.)

gram or so either greater or less than 10.0000 g. Even if the manufac-

turer did find that he had made a perfect 10.0000 g. piece, as checked on

his balance, this same weight on another balance might not act as a per-

fect ten. This is due, chiefly, to the fact that in spite of the best efforts of

the manufacturer the arms of the balance may be slightly unequal in

length, giving slightly different beam leverages to the right and left of the

fulcrum. It is necessary then, for really accurate weighing, to check each

piece in the box of weights against a standard weight and, once this has

been done, always to use these same weights with a given balance. The

calibration of weights will be taken up in a later section.

Two classes of weights may be purchased: Class S weights for weigh-

ings in ordinary analyses, and Class M weights for work of high accuracy.

Once calibrated, however, either class is entirely satisfactory. Table 1

lists the tolerances recommended by the U.S. Bureau of Standards.

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Table i



Before taking up different methods of weighing, the following pointsshould be studied carefully until so familiar that they become second

nature to the analyst.

1. Always use the same balance throughout a given analysis. If the

balance is shared by two or more persons, cooperate with the others in

keeping the balance in perfect order.

2. The balance must be mounted so that it is free from vibration. In

laboratories in which brick walls of the balance room extend through floors

continuously to the earth the balance may be placed upon a shelf attached

to the wall (the shelf must never rest upon the floor) ; otherwise a pier rest-

ing upon the ground should be employed.3. The balance must never be subjected to direct sunlight or to drafts.

It is best to keep the windows of the room closed and the shades drawn.The room should be well lighted with several lights so that no shadows are


4. Avoid the habit of comfortably placing the elbow alongside the

balance with the hand against the side of the balance. This heats up one

side of the balance and makes accurate weighings impossible.5. Never weigh an object which is warm. Wait until it has reached

room temperature. A hot object produces an upward motion of the air

around the object and this causes low results in weighing.6. Except when weighing an object which cannot possibly injure the

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pan, never place the substance directly upon the pan; instead place it in

a container: a watch glass, crucible, etc. It is unnecessary to place anydesiccant within the balance case. 1

7. Usually the first step in a weighing is to obtain the equilibrium

point (also called the rest point and, if the pans are empty, the zero point),

that is, the position over the scale at which the pointer freely rests. If the

equilibrium point is more than one scale division from the center of the

scale, request the instructor to readjust it. This he may do by varying the

mass on one side of the beam through the adjusting screw.

8. When the beam is released bring the agate knife-edges into con-

tact with their planes very carefully; abrupt lowering of the beam mightresult in chipping the agate prism. Put the beam in motion by momen-

tarily dropping the rider at any point on the beam. This is better than

wafting a current of air toward the pan with the hand, which sometimes

results in accidentally hitting the pan and also may blow away any

powdered substance being weighed.

9. During a weighing, when it is desired to stop the oscillation of the

pointer, do so by releasing the pan arrests at the instant the pointer

crosses the mid-point of the scale. This prevents jarring. Then carefully

raise the beam support. In putting weights on the pan or removing them

be sure that the beam is locked.

10. Weights should be handled only with the forceps. Weights are

always placed on the right-hand pan. Large weights should be placed near

the center of the pan, for otherwise the pan will not hang in a plumb posi-

tion when the arrests are lowered. Place fractional weights on the pan in

order of their denominations. In handling large weights hold the forceps

with the concave edge up. With fractional weights hold the concave edgedown and pick up the weights with the points of the forceps.

11. Keep the fractional weights in the box in a definite sequence (see

Fig. 16). It will save time if the turned-up edges are arranged consistently,

e.g., always at the top of their compartments.12. In recording the sum of the weights in the notebook first list the

weights which are missing from the box. Then check the list as the weightsare removed from the pan. The checking requires only a moment of

time; one error in summing up the weights may cost days or even a week

in repeating an experiment.13. When a weighing has been completed see that everything is left

in good order: the beam support and pan arrests are up; the rider is on its

hook and the rod has been pushed back into the case; the base of the

balance and the pans are brushed clean; the window is closed; the chair

has not been left out on the floor where someone may trip over it.

Equilibrium Point. The equilibrium point or rest point (the zero

point if the pans are empty) of the balance is that position of the pointeri Kuhn, Chem. Ztg., 34, 1097 (1910).

Page 65: quimica inorganica cuantitativa


on the scale at which it eventually would come to rest after being set into

motion. It is not necessary to wait until the pointer actually comes to a

standstill. Rather, if the points to which it successively swings to one side

and the other are noted the arithmetic mean will give the equilibrium

point, provided only that an odd number of readings are taken on one

side and an even number on the other side. Due to friction within the

balance, and to air resistance, the oscillations gradually become of smaller

and smaller amplitude, and each swing accordingly will become slightly

shorter than the preceding one. Thus the final position of the pointer will

not be midway between the extremities of two successive swings, nor

even midway between the average of two successive positions on the

right and two on the left. The equilibrium point will be, however, that

FIG. 17. The scale of an analytical balance.

position halfway between the average of n points on one side and n + J

on the other, successively taken.

Suppose, as a simple example, a balance is so adjusted that the

equilibrium point is actually 10.0, the scale being numbered as seen in

Fig. 17. (This does away with negative values which result if the middle of

the scale is numbered zero.) Suppose further that the slow-down amountsto one scale division for each half oscillation. (As a matter of fact the de-

crease is seldom this great.) If the first reading were at 3.0 the succeeding

readings would be as follows:


1st reading 3.0

2nd reading 5.0

3rd reading 7.0


1st reading 16.0

2nd reading 14.0

3rd reading 12.0

If we took the first two readings both right and left we would have

averages of 4.0 on the left and 15.0 on the right. The mean of 4.0 and 15.0

is 9.5, which differs by 0.5 from the equilibrium point of 10.0 assumed in

our example. Similarly using all three values on both sides gives 9.5.

However, if the first two readings on the left and the first alone on the

right, or if all three on the left and the first two on the right, are used, the

average of 10.0 is obtained; we repeat, then, that an odd number of read-

ings must be taken on one side and an even number on the other, and all

must be taken successively.

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To determine the equilibrium point proceed as follows. Brush off the

pans with a camel's-hair brush, then close the window and lower the beam

support and the pan arrests. Drop the rider momentarily on the beam,

remove the rider, and allow one or two oscillations to take place. An

amplitude of about 8 to 12 divisions is suitable. Then record the readings

of the pointer, twice on one side and three times on the other. Since the

scale is marked off in units readings are taken to the tenths, for this gives

one doubtful digit. To illustrate:

Left Right

1st reading 5.1 1st reading 15.0

2nd reading 5.3 2nd reading 14.8

3rd reading 5.5 Average 14.9

Average 5.3 _J

2| 20.2

Equilibrium point 10.1



FIG. 18. Factors influencing the sensitivity of a balance.


The sensitivity of the balance is defined as the number of scale di-

visions by which the equilibrium point of the balance shifts due to an

excess of one unit of weight, usually taken as 1 mg., on one of the pans.2

Why should different balances have different sensitivities, and whyshould the sensitivity of a given balance, instead of being a constant,

change with different loads? The answers to these questions are found in

the laws governing simple levers. A lever is in a condition of equilibrium

when the force of moments which tend to impart a clockwise rotation

equals that tending to give a counterclockwise rotation. In Fig. 18 let the

2Sensitivity is sometimes defined in other ways. Manufacturers refer to the sensi-

tivity as the smallest mass which produces an observable shift of the equilibrium point.

The term also occasionally is used to denote the number of tenths of milligrams which

will cause a shift of the equilibrium point amounting to one scale division.

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solid lines represent the equilibrium position of the balance with no load

and the dotted lines the equilibrium position after the excess weight, m,has been added to the right-hand pan. Two opposing forces are at play:the force tending to cause a clockwise rotation produced by the effect of

gravity, g, on the excess weight, and the force tending to cause the counter-

clockwise rotation produced by the effect of gravity on the mass of the

beam, M, acting through its center of gravity, B, which is located at a dis-

tance, d, from the fulcrum, 0.

The effect of a force in turning a body about any axis is called the

"moment" of the force about that axis; it is calculated as the product of

the force (acceleration of gravity, g, multiplied by the mass) and the

perpendicular distance from the axis to the line of action of the force.


(1) M0.BC = m^.OA

Now angle BOG is equal to angle AOA' since the corresponding sides of

the two triangles are mutually perpendicular. Call these angles a. Fur-

thermore, line OB = rfand OA' = L, the arm of the beam. Thus

BC = d sin a

and OA = L cos a

Substituting these values in equation (1) yields

sin ot = mglt cos asin a A mL

With the balance we are concerned only with small angles in which case

the tangent is practically equal to the angle itself (in radians). It follows

therefore that the sensitivity of the balance, S, i.e., the displacement or

shift made by the pointer,

/o\ c ^ _ LW/ ^ :==' T|/r~fv m Mrf

From equation (3) it is seen that the angular displacement, and thus

the sensitivity, increases as the length of the beam increases. It decreases

as the mass of the beam increases; thus the beam and the supported pansshould be of light construction, but rigidity of the beam should not be

sacrificed. The sensitivity also decreases as the distance from the center

of gravity becomes greater. Therefore the sensitivity may be increased by

raising the center of gravity of the balance (by elevating the gravity bob;see Fig. 11). The center of gravity, however, can never be above the

knife-edges, since then the balance would be in unstable equilibrium and

could not oscillate. With a higher center of gravity and greater sensitivity

Page 68: quimica inorganica cuantitativa


the period of oscillation becomes greater, and since the latter extends the

time necessary to make a weighing, a compromise between higher sensi-

tivity and shorter period must be made.

The assumption has been made that the three knife-edges of the bal-

ance are in the same plane. However, with heavier loads on the balance

the beam will bend slightly, with the result that the knife-edges at the

terminals are below the one at the fulcrum ; this may bring about a rapid

falling off of the sensitivity with increasing loads. In fact some balances

which are to be used regularly for rather heavy loads are constructed

purposely with the extreme knife-edges slightly higher than that at the

center, so that under loads the three come more nearly within the same



As will be explained later, for accurate weighings the figure for the

fourth decimal place is established by use of the sensitivity of the balance.

The sensitivity at various loads therefore must be determined. To do so

proceed as follows: Determine the equilibrium point with the pans emptyas already explained. Do this in duplicate. The duplicate runs should

check within 0.2 of a scale division. Suppose the mean result is 9.8, that

is, two-tenths of a scale division to the left of the center-point. Now place

the rider momentarily at any point on the beam so that the pointer be-

gins to swing with a moderate amplitude. Then place the rider exactly

on the 1 mg. mark on the beam. The balance now operates with one more

milligram of weight on the right side than on the left; the equilibrium

point therefore must shift to the left and the number of scale divisions bywhich it does shift will, by definition, be the sensitivity of the balance for a

zero load. Determine this new equilibrium point in duplicate, again being

sure that these duplicate results check within 0.2 scale division. Suppose

the mean result is 6.5. The difference between this and the original equi-

librium point (before adding the 1 mg. excess weight) is 9.8 6.5 = 3.3.

This then is the sensitivity of the balance for a zero load. Let us repeat

that this simply means that when the two pans are in a state of imbalance

to the extent of exactly 1 mg., the equilibrium point shifts by 3.3 scale

divisions. Conversely, if later, in making a weighing with this balance, we

found that the final equilibrium point differed from the original equi-

librium point by 3.3 scale divisions, the object being weighed would differ

in weight from the sum of weights being used by 1 mg.Next determine the sensitivity of the balance for a 10 g. load. To do

this, place a 10 g. weight on each of the pans (the rider is not on the beam,

of course) and determine the equilibrium point, again in duplicate. Do not

be disturbed if this equilibrium point is not the same as was obtained with

both pans empty. Remember that the beam may be slightly bent under

Page 69: quimica inorganica cuantitativa


the 10 g. load, that friction will be greater under the heavier load, and,

furthermore, that few weights are perfect so that the two 10 g. pieces may,in fact, be of not quite the same mass. For any of these reasons the equi-librium point now may not be at 9.8. Say that it proves to be 10.1. Next,after setting the pointer to swinging, place the rider at 1 mg. and deter-

mine again the equilibrium point. If this were, say, 7.0, the sensitivity for

the 10 g. load obviously would be 10.1 7.0 = 3.1. In like mannerdetermine the sensitivity for a 20 g., 30 g., 40 g., and 50 g. load, and then

make a graph of the results, plotting the sensitivity as the ordinate andthe load in grains as the abscissa. Place the graph, which will be similar to




10 20 30 40 50Load in g.

FIG. 19. Sensitivity curve of a balance.

Fig. 19, in the upper left-hand corner of the balance window. It should be

mentioned that a balance should not be subjected to a load for which the

sensitivity is less than 40 per cent of the maximum sensitivity.

It sometimes happens in determining the sensitivity of the balance,

especially for heavier loads, that the sum of the (imperfect) weights on

one pan differs from the sum on the other pan (even though the same

denominations are placed in both pans) to such an extent that the pointer

swings entirely off the scale. If this happens place the rider at such a

position on the beam that the initial equilibrium point for the load is

near 10. After determining the equilibrium point in duplicate move the

location of the rider by exactly 1 mg. and determine the new equilibrium

point in duplicate. This will give the sensitivity even though the rider was

used twice for the given load, for here also the shift in equilibrium pointwas caused by a difference in weights of 1 mg. by the rider.


Direct Weighing. Method of Swings. The object to be weighed

always is placed on the left-hand pan. Due to inequality in the lengths of

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the arms of the balance, weighings made by direct methods do not repre-

sent the true masses of the objects but are proportional to them ; thus they

give a correct ratio of masses. This method of weighing is reliable when

results are sought on a percentage basis, since any difference in the lengths

of the two arms is compensated. (If the absolute or true value of the mass

is wanted the weighing may be carried out by the direct method and the

weight obtained multiplied by the arm ratio, p. 54, or else either the Gauss

method, p. 54, or the Borda method, p. 55, of weighing may be employed.)

Procedure. First determine the equilibrium point of the balance.

Then place the object on the left-hand pan. Next with the forceps place on

the right-hand pan that integral weight judged to be nearest the mass of

the object. Lower the beam support just enough to reveal in which direc-

tion the pointer moves. If the weight chosen is too heavy, lock the beam

support and substitute the next lighter weight. If this weight is not too

heavy it is left on the pan and the next weight in descending order is

added, and the beam support again is lowered to show the inequality. If

the total weight at this point is too great the last piece added is replaced

with the one just lighter than it; if the total weight is not too heavy the

last weight which was added is left on the pan and, in addition, the next

lower weight is placed on the pan. This process is continued with smaller

and smaller integral weights until finally the 1 g. weight is either used or

rejected. Then the beam support is kept lowered as the fractional gram

weights are tried, since the pan arrests are strong enough to hold the

pointer midway on the scale when the state of imbalance is less than a

gram. The fractional weights are added in the same systematic fashion as

were the integral weights starting with the 500 mg. piece. The state of

imbalance may be tested by lowering the pan arrests just enough to show

which direction the pointer takes. In the above manner the total number

of separate pieces used will automatically be the smallest number possible.

When two different pieces of a given denomination are available always

use the one-dot piece in preference to the two-dot weight. (See p. 57.)

The rider is now placed upon the beam and adjusted so as to cause

the same equilibrium point as was found at the beginning with no load.

By this method the rider most likely will not be situated finally at a whole

milligram division on the beam, and, whether this is so or not, the last

two digits i.e., for the third and fourth decimal places are read directly

from the beam.

The figure for the fourth decimal place is more quickly established in

the following manner: Instead of placing the rider at such a point on the

beam as will regain the original equilibrium point, simply locate it on that

whole milligram division on the beam which is most nearly correct to

balance the object being weighed. Then determine in duplicate the equi-

librium point. The difference between this equilibrium point and that

for empty pans is divided by the sensitivity of the balance (for the par-

Page 71: quimica inorganica cuantitativa


ticular load; consult the sensitivity curve). The quotient gives the degree

of imbalance in terms of milligrams, and this must be added to or sub-

tracted from the sum of the weights on the right pan, including the rider.

The following example should make the method clear.

(a) Equilibrium point, zero load 9.8

(b) Equilibrium point with weights

totaling 9.87 on pan, and rider

at 6.0 on the beam 10.9

(c) Shift of equilibrium point 1.1 scale divisions

(d) Sensitivity of balance for 10 g.

load 3.1

(e) State of imbalance = 1.1/3.1 = 0.3 mg.

The total weight therefore is recorded as 9.876 g. + 0.3 mg. = 9.8763 g.

This is still not the weight accepted by the analyst, however, since the

weights correction (see Table 3, p. 61) has not yet been applied. Note that

this sensitivity correction of 0.3 mg. is added since the equilibrium pointwith the load was greater than that with pans empty, meaning, of course,

that the 9.876 g. was too light for the object. When the equilibrium pointwith the load is smaller than that with pans empty the sensitivity cor-

rection is subtracted from the total weights on the pan. It hardly needs

to be mentioned that to attain a precision of a tenth milligram the sensi-

tivity of the balance must be 2.0 or greater since a weighing involves

taking two equilibrium points both subject to an error of 0.1 scale division.

Direct Weighing by Single Deflection Method. A more rapid pro-cedure known as the single deflection method3

is often employed. It is not

recommended when weighings of the highest accuracy are desired 4 nor

when hygroscopic substances are weighed. By the adjustment screw of

the beam the left arm of the balance is made permanently heavier than

the other so that when the pointer is free to swing, a definite deflection

results due to the added moment on the left arm. The deflection should

amount to about five or six scale divisions, i.e., to about 15 or 16 on the

scale. In weighing an object it is placed on the left pan as usual and

weights are placed on the right pan, and the rider located so as to givethe same single deflection as with pans empty. Or, provided a sensitivity

curve has been constructed also using the single deflection method, the

duplication of the original equilibrium point is unnecessary; rather, the

fourth decimal place is calculated from the two equilibrium points andthe sensitivity as usual.

The only advantage of this method of weighing is its speed. Its use

requires a balance with separate beam and pan releases and the latter

must be carefully adjusted so that it gives no impetus to the pans when

Britton, J. Am. Chem. Soc., 41, 1151 (1919).4Lin, J. Chem. Education, 16, 340 (1939).

Page 72: quimica inorganica cuantitativa


released. If the single deflection method is to be used the beginner should

practice releasing the pans until a constant deflection can be obtained at


Method of Double Weighing (Gauss). For very accurate work this

method, proposed by Gauss, is often employed. It eliminates errors other-

wise resulting from inequality in length of balance arms, and the results

give about twice the accuracy obtained by other methods. The object is

weighed first on one pan and then on the other. The mean of the two re-

sults is taken as a measure of the true weight.

It follows from the principle of the lever that if an object of true mass,

M, is placed upon the left-hand pan and is counterbalanced by weights,

W r , on the right-hand pan,

(4) ML, = WrLr

where Lt and Lr are the lengths of the left and right arms of the balance,

respectively. Placing the object now on the right-hand pan and counter-

balancing with weights Wi on the left-hand pan, we have

(5) ML, = W.L,

The product of equations (4) and (5) yields

M 2 = WtWr

from which

Usually the values Wi and Wr are nearly equal and as a close approxima-tion

(6) M . 2L+W.

M, determined by this method, gives the true mass of the object.

Ratio of Lengths of Balance Arms. Equations (4) and (5) above

may be combined and used to calculate the ratio of the arms of the bal-

ance. Dividing (4) by (5) yields


In a good balance the arms will seldom differ in length by more than 1

part in 50,000. It must be realized that the ratio L,/Lr is not a constant

even for a given balance but is a variable dependent upon conditions.

If one arm is at a slightly higher temperature than the other the ratio

would be changed. Under different loads the ratio may be different. It

follows that the ratio of arm lengths as determined under one set of

Page 73: quimica inorganica cuantitativa


conditions must not be used for the calculation of the true mass of an

object when weighed under different conditions.

Weighing by Substitution (Borda). This is a second method used

for finding the true mass of an object regardless of the fact that the bal-

ance arms may be of slightly different lengths. The object of mass, M, is

placed on one pan, say the left-hand pan, and counterbalanced by another

body (e.g., copper shot) of weight T, called a tare. The equilibrium point

is then determined. The object is next removed from the pan and re-

placed by calibrated Weights (see p. 57) of such mass, W, as to give the

same equilibrium point. From the first counterbalancing we have

(8) ML, = TL,

and from the second

(9) WL! = TLr

Therefore ML, =or M = WThis is to say, ignoring the buoyancy effect of the air which will be dis-

cussed below, the mass of the object is equal to that of the weights; this

must be true since both objects and weights have been acting through the

leverage of the same arm, LI.

Whenever highest accuracy is essential, one of the methods for obtain-

ing true weights must be employed. For example, in the calibration of

weights and of volumetric apparatus (p. 57 and Chapter 5) the substitu-

tion method has been recommended.

Buoyancy Correction in Weighing. It will be recalled that Archi-

medes' Principle states that when a body is completely immersed in a

fluid, liquid or gaseous, its weight is diminished by an amount equal to

the weight of the fluid displaced. When a weighing is made in air both the

object and the weights are subject to the buoyancy effect of the surround-

ing air. The denser the body is, the less air it will displace per unit of

mass. The weight of an object in air is the result of the do\v award pull of

gravity and the upward buoyant effect of the air upon both object and

weights. The net effect of the latter force will depend upon which dis-

places more air, the object or the weights; since the object weighed is

almost always less dense than the weights themselves it follows that the

weight of the object in air usually will be less than its weight in vacuo. The

buoyancy effect will be zero only in the rare instance when the object anjthe weights displace the same amount of air, that is, when their densjare equal. Because it is impracticable to carry out a weighing i

a correction for buoyancy always must be made in very exact

rule the correction is ignored for weighings in quantitative

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because the air displaced by the object is small and also because there is

generally a canceling out of the effect due to the fact that the calculations

so often are based upon two weighings (the sample and the precipitate,

for example) involving substances of not widely differing densities. If,

however, large volumes of air are displaced by the object, as in the calibra-

tion of volumetric apparatus, or in the weighing of gases and in density

determinations, the buoyancy effect must be taken into account.

The weight in vacua may be calculated as follows. It is obvious that the

weight in vacua, Wv , is the weight in air, Wa , plus the buoyancy effect onthe object, minus the buoyancy effect on the weights, or

(10) Wv= Wa + (Buoyancy Effect, Object)

(Buoyancy Effect, Weights)

Now the buoyancy effect on the object is the product of the volume of the

air displaced by the object (equal to the volume of the object itself) andthe density of the air. Thus the first term within the parentheses of equa-

Wtion (10) becomes

^~Da , where \VV is the in vacuo weight of the object,

Da and D are respectively the densities of air and of the object in g. per

ml. Similarly the second term in the parentheses becomes^ Da , whereDw

Wv is the in vacua value of the weights employed and Dw is the density of

the weights. Thus equation (10) becomes

(11) W. = W. + ' _ WJtt.

DO iJw

In general Wv and Wa do not greatly differ, so that equation (11) may bewritten

W ~ w , W.Da W.D.Wv = Wa + ~

or Wv ^ V

from which

(12) Wv S ^

The density of air varies with the temperature, the barometric pressure andthe humidity. In the laboratory the air is usually moist arid between tem-

pe^atures of 15 and 30 and pressures of 720 mm. and 780 mm. A milliliter

of (fry air at 15 and 780 mm. weighs 0.00126 g., whereas a milliliter of air

Saturated with water vapor at 30 and 720 mm. weighs 0.00108 g. As an

average between these extremes we may employ a density for moist air of

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0.0012 and use this value in equation (12). Doing this, and substituting a

value of 8.4 as the density of brass, for Dw , we have

(13) W. ~ W.[l+ 0.0012


for calculating the true, in vacuo, weight of objects weighed in air with

brass weights at ordinary room conditions.


It already has been pointed out that the weights in any set are not

likely to have exactly the mass that their nominal values indicate, and

therefore should be calibrated. For the ordinary weighings of quantita-

tive analysis it is not necessary that we know the absolute values of the

weights. It is essential, however, that the various weights of a given set

agree among themselves; i.e., either the 50 mg. weight must be exactly

5 times as heavy as the 10 mg. weight or else we must know to whatextent it varies from 5 times the 10 mg. weight; either the 2 g. weightmust be exactly 200 times as heavy as the 10 mg. weight or else wemust know to what extent it departs from 200 times the 10 mg. weight,and so on. The calibration of the weights consists of the determination

of such deviations. The calibration may be relative, in which case wedetermine the values of the several denominations based upon one

single weight of the set; or it may be absolute, in which case the value

of each weight in the set is based upon a single certified weight the

absolute value of which has been established by the U.S. Bureau of


Before beginning a calibration the weights should be wiped clean

with a piece of Kleenex tissue. If the set contains more than one pieceof a given denomination, e.g., two 100 mg. weights, the duplicatesshould be given distinguishing marks. Usually they are marked with

one and two dots by the instructor. We shall designate such duplicatesas 100', 100", etc. An auxiliary set of cheap weights must be available

to be used as tares.

The principle of a calibration is as follows. Either the 5 mg. weight

(for balances having a beam divided into less than 10 mg. divisions) or

the 10' mg. weight (for balances having a 10 mg. beam) is arbitrarily

assigned a weight of 5.00 (or 10.00) units. The unit is approximately,but only by coincidence exactly, 1 mg. Suppose we are working with a

balance with a 10 mg. rider; i.e., there are 10 divisions on the beam.

First, by the substitution method, the rider is compared with the

6Richards, J. Am. Chem. Soc., 22, 144 (1900); Semon, J. Chem. Education, 2, 132

(1925); Hurley, Ind. Eng. Chem., Anal Ed., 9, 239 (1937); Thornton, J. Chem. Edu-cation, 14, 270 (1937); Blade, Ind. Eng. Chem., Anal. Ed., 11, 499 (1939).

Page 76: quimica inorganica cuantitativa


10' mg. weight; then the 10" mg. weight is compared with the 10' mg.

weight; the 20 mg. weight is compared with the 10' mg. plus the

10" mg. pieces, and so on through all the fractional weights and the

integral weights. These comparisons yield the values for each piece in

relative terms, of course, based upon our arbitrarily assigned unit. As

we proceed to whole-gram weights the comparative value for a given

weight differs rather widely from the nominal value of the weight.

Accordingly, it is customary, once all the relative values are obtained,

to assign an exact value to one of the larger weights, usually 10.0000 g.

to the 10 g. weight. It then becomes the new (arbitrary) standard in

terms of which all of the other pieces are expressed. Finally, if a certified

weight is at hand, say a 10 g. unit checked by the Bureau of Standards,

the absolute values of all the weights of the set may be computed.Procedure. First calibrate the rider. Place the rider exactly on the 10

rng. division; put an auxiliary 10 mg. weight on the left-hand pan as a

tare and determine the equilibrium point in duplicate, calculations being

made to the hundredths. Remove the rider from the beam, place the

10' mg. weight on the right-hand pan and again determine the equi-

librium point. To illustrate the calculation involved, assume the two

equilibrium points were 9.84 and 9.67 respectively, and that the sensi-

tivity of the balance was 2.8. Then -^s~<r~

= 0.06, which is theZ.o

amount by which the 10' mg. weight and the rider differ. Since the

second equilibrium point is smaller than the first it is obvious that

the rider is lighter than the 10' mg. weight, and since we assigned a value

of 10.00 units to the 10' mg. weight the rider must weigh 9.94 units.

Now compare the 10" rng. weight with the 10' mg. weight in the

same general manner. Suppose it is found to be 0.08 unit lighter than

the standard 10;

mg. weight; it then would have a value of 9.92 units.

Next calibrate the 20 mg. weight. To do this, place the two 10 mg.

weights on the right-hand pan and a 20 mg. auxiliary weight on the

left-hand pan. Determine the equilibrium point. Remove the two 10

mg. weights from the right-hand pan, replace with the 20 mg. weightfrom the analytical set and determine the new equilibrium point. Again,

just as before, calculate the number of units by which the 20 mg. weightfrom the analytical set is heavier or lighter than the sum of the two10 mg. weights of the set. Say the 20 mg. weight is 0.06 unit heavier

than the sum of the two tens. Then the 20 mg. weight must have a

value of (10.00 + 9.92) + 0.06 = 19.98 units.

The value of the 50 mg. weight is next determined in terms of the

20 rng. plus the two 10 mg. weights plus the rider placed at the 10 mg.division of the beam, again, of course, by substitution, using an auxiliary50 mg. weight as a tare. In the same manner comparisons are made for

Page 77: quimica inorganica cuantitativa


each denomination of the whole analytical set, both fractional and

integral weights. When this is completed, data such as that given in

the first five columns of Table 2 may be assembled.

All of the figures of the table through Column 5 are based upon the

assignment of 10.00 units to the 10' mg. weight. The choice of such a

small weight as the standard is advantageous in that one unit is just

about 1 mg., but this would result in rather large corrections as wecome to the larger denominations (see Column 5). We may circumvent

this inconvenience by assigning a new, heavier standard and reevalu-

ating all denominations on the new basis. For example the 10' g. weight

may now be assumed exactly equal to 10.0000 g. Upon this basis, since

we have established the relative value of the 10' g. weight as 9941.20

units (see Table 2),

9941.20 units = 10.00000 g.

Therefore 1 unit =-^941 20 g *

= 1 -005905 x 10~3


If then we multiply each unit value of Column 5 by this value of one

unit, 1.005905 X 10~3g., we shall have the relative value of each de-

nomination in terms of grains. The same thing may be accomplished,with less labor, in the following manner: Calculate the aliquot part of

9941.20 for each weight by multiplying by that part of 10 which the

denomination represents. That is to say, for the 2 g. weight, multiply9941.20 by 0.2; for the 50 g. weight multiply 9941.20 by 5, etc. Place

these aliquot parts in Column 6. If a given number in Column 5 is

identical with the corresponding number in Column 6 then, and only

then, that piece is a perfect weight (as based upon our arbitrary assign-

ment). If a number in Column 5 is smaller (or larger) than the corre-

sponding number in Column 6 then that piece is not a perfect weightbut is lighter (or heavier) by an amount represented by the difference

in the two columns. Column 5 minus Column 6 thus gives the correction,

in milligrams, which must be applied to the denominational value of

each weight in order to have its correct value, based on the assumptionthat the 10' g. weight is perfect. Place these corrections in Column 7

and then obtain Column 8 by algebraically adding the correction to the

figures stamped on each piece. In Column 8, therefore, the true, relative

value of each weight is found.

It might be repeated for emphasis that in most analytical work it is

the ratio of two weighings which is required. However, if an absolute

calibration is desired it is only necessary to compare, by the substitution

method, say the 10' g. weight with a 10 g. weight the absolute value of

which has been certified by the U.S. Bureau of Standards. From the

comparison the absolute value of the 10' g. weight of the analytical

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i 18









+ 1 + 1 1 + + + \ + + + \ \ +


eo-ioor-^l '

OOOOOOOOOi-HOOOOOOO1 1 + 1 1 +4- 1 1 1 + 1 1 1 1 +


'd ^a &


e e o e o H ui b b e o

Page 79: quimica inorganica cuantitativa


set is calculated. Suppose in this manner the 10' g. weight is found to

have an absolute value of 10.00004 g. Obviously the absolute correction

for the 10' g. weight is + 0.04 mg. The correction for the 5 g. weight will be

^nrrA^o /9941.20 - 0.04\ IA ,- j *u u i . i *4970.73 (

J= +-l5 mS-> and the absolute value of

the 5 g. weight is 5.00015 g. In this way the absolute weights of the

entire set may be calculated. (Absolute values have not been included in

Table 2.)

Calibration of Weights by a Complete, Certified Set. Some labo-

ratories possess a complete set of weights which have been certified bythe Bureau of Standards. A certificate is furnished showing the true

weight of every piece, [f such a certified set is available each piece from

the analytical set may be checked against the corresponding certified

piece by the method of substitution. In this way we avoid the labor of

calculating the values of all pieces on the basis of the arbitrarily assigned

standard and the consequent possibility of accumulated errors. Further-

more, the corrections are at once on an absolute basis.


To facilitate making the proper weights correction after weighing an

object, it is well to prepare a table which, at a glance, will give the

proper correction to apply. Two points in particular must be borne in

mind in assembling the data for such a table. It is presumed that the

weighing has been made in accordance with the rule laid down in the

procedure for direct weighing (p. 52) and that therefore the smallest

possible total number of pieces has been used. This is important be-

cause, for example, the weights correction for an object of 5 g. is different

if the weighing is done with the single 5 g. weight from what it would

be were the weighing carried out using two 2 g. weights and the 1 g.

weight. Remember too that a "one-dot" weight is always selected in

preference to a "two-dot" piece of the same denomination. In con-

structing the table, then, we follow the rule of avoiding an unneces-

sarily large number of weights, and of using the proper piece when a

choice is possible.

Another point to keep in mind is that the correction for each denomi-

nation in Table 3 has been obtained to the hundredth milligram. In

accord with the rules for significant figures, even though the table of

corrections will list corrections only to the tenth milligram, the roundingoff should be deferred until the algebraic summation has been made. Toillustrate: say the individual corrections, in milligrams, for the 100' mg.,

the 50 mg. and the 10' mg. weights are, respectively, 0.10, 0.01 and

+0.06. Should these corrections first be rounded off to the tenths and

Page 80: quimica inorganica cuantitativa


Table 3



then the algebraic addition made, the correction for a total weight of

160 mg. would be -0.1 - 0.0 + 0.1 = 0.0. However, if the addition is

made first and the sum then is rounded off to the tenths, the resulting

correction for 160 mg. is -0.10 - 0.01 + 0.06 = -0.05, or, rounded off,

0.1, which is the proper correction. Table 3 is the weights-correction

table prepared from data taken from Table 2.


What has been presented in the foregoing sections emphasizes the

fact that there are many sources of possible error whenever a weighing

is made. The significance of several of them has already been evaluated;

for example, that due to inequality of length of balance arms (pp. 54, 55),

that due to buoyancy of the air (p. 55) and that due to inaccuracy of

the weights (p. 57). Means of overcoming or minimizing such errors

have been discussed. In addition, errors will result from weighing objects

not at room temperature, or if uneven temperature prevails within the


Page 81: quimica inorganica cuantitativa


Still another cause for poor results comes from wiping the object,

especially if made of glass or of porcelain, with a cloth, thereby impartinga static charge of electricity to the object. When this happens erratic

results in the determination of the equilibrium point with the load are

immediately noticed; in such cases the weighing cannot be completeduntil the electrical charge is dissipated, which may take several minutes.

It often happens that a substance being weighed becomes heavier

even while the weighing is being carried out. This means that somethingis being taken up by the substance from the air. Hygroscopic materials

take up water, and certain substances like calcium oxide take up carbon

dioxide, with sufficient rapidity to make reproducible equilibrium points

impossible. When this occurs it is best to place the approximately correct

amount of weights upon the right pan before removing the object from

the desiccator. When this has been done the object is placed on the left

pan, the rider quickly adjusted and the equilibrium point determined

at once.



In weighing a sample of a powdered material for analysis, either of

two methods may be employed. Balanced watch glasses (pairs, which

are of almost equal weight) may be used. With balanced watch glasses

one does not need the equilibrium point with the pans empty. Rather

the equilibrium point with the watch glasses in the pans is obtained.

The weights (equal to the desired weight of the sample) are placed in the

right watch glass and the powdered substance is added from a spoon to

the left glass until slightly in excess of the weights. The exact weightthen is determined as usual through proper placement of the rider and

the determination of the equilibrium point with the load.

Powdered substances also may be weighed out by difference. This

method, moreover, is useful especially in weighing sublimable or hygro-

scopic samples for analysis that is, when the substance is likely to lose

or to gain weight while being weighed. Several grams of the material

are placed in a vial which is stoppered and weighed. Then holding the

vial with cork-tipped forceps, the analyst removes the stopper with

tongs and gently taps out a sample of appropriate size. The stopper is

replaced and the vial again weighed. The decrease in weight represents

the amount of sample obtained.

Questions and Problems

1. Name several factors which affect the sensitivity of the balance. Explain.

2. Explain how the sensitivity of a balance may be changed.

3. What objections are there to overloading a balance?

Page 82: quimica inorganica cuantitativa


4. What advantages have the method of double-weighing and the substitution

method over the direct method of weighing?

5. Under what circumstances is the buoyancy correction unimportant in


6. In weighing by the method of substitution should the tare (i.e., counterpoise)be of about the same density as that of the weights used? Why or why not?

7. An object when placed on the left-hand pan of a balance has an apparent

weight of 18.2468 g. When transferred to the right-hand pan and weighed, the

apparent weight is 18.2486 g. (a).What is the true weight of the object? (b)

What is the ratio of the lengths of the balance arms? (c) In parts per million,

how much do the lengths of the arms vary?Answer: (a) 18.2477; (b) 1.00000:1.00005; (c) 50.

8. A sample of a certain alloy weighs 56.8873 g. in air. Suspended by a wire and

weighed in water the alloy weighs 49.1960 g. The wire immersed in water to

the same depth weighs 0.3210 g. All weighings were made with brass weights.

(a) Calculate the density of the alloy, (b) Calculate the true weight, in vacuo,

of the alloy.

Answer: (a) D = 7.1; (b) 56.8888 g.

9. Exactly 1.0000 g. samples of substances with densities of 0.25, 0.50, 1.00,

2.00 and 8.00 are weighed in air with brass weights. Calculate the differences

between the weights in air and the corresponding weights in vacuo. Make a

graph of these differences plotted against the densities.

10. A piece of platinum having a density of 21.37 weighs 5.0000 g. when weighedin air with a certain set of weights. If the platinum weighs 4.9981 g. in vacuo,

what must be the density of the weights which were used?

Answer: 2.8.

11. A platinum dish weighs 28.7654 g. in air when weighed with aluminum

weights. Calculate its weight in vacuo.

Answer: 28.7542 g.

12. The weight of a platinum object weighed in air with aluminum weights is

greater than its weight in vacuo. The weight of an aluminum object weighedin air with platinum weights is less than its weight in vacuo. The weight of a

brass object weighed in air with brass weights is the same as its weight in

vacuo. Explain these statements.

Page 83: quimica inorganica cuantitativa

Chapter 5


JT RECISE measurements of Volume, usually of liquid, are made by the

analyst almost as often as those of mass. The more commonly employedinstruments for measuring volumes are the buret, the pipet and the

volumetric flask. The first two usually are used to deliver a definite

quantity of liquid while the last contains a given volume. Kach is marked

with etched lines by the manufacturer to designate the volume delivered

by or contained in the apparatus. Just as in the case of the weights used

with the analytical balance, it is probable that the number appearing

on the volumetric flask, for example, is somewhat in error. If the flask is

marked 500 ml. the probability is that it holds a little more or a little

less than 500 ml. Therefore for the best work all such apparatus should

be calibrated; that is, we ourselves should determine the capacity of the

instrument as carefully as possible. Otherwise, every time we use the

apparatus we introduce an error into our work, and frequently such

errors exceed the errors inherent in the analysis.

The unit of volume is the liter (1.). The liter is that volume occupied

by 1 kilogram of water at the temperature at which the density of water

is greatest, namely, 3.98, and at a pressure of 1 atmosphere. One-

thousandth of the liter is a milliliter (ml.). The millilitcr is almost but

not quite identical with the cubic centimeter (cc.) which, of course, is the

volume of a cube having an edge 1 centimeter in length. Originally it

was intended that the two would be the same, but because of small

experimental errors a difference between the two exists amounting to

28 parts per million, or 0.028 parts in 1000. (Prove this from the fact

that exactly 1 ml. = 1.000028 cc.) This variation is so slight as to be

insignificant in volumetric analyses where the expected precision usu-

ally is 1 part per 1000. Since the milliliter is the unit employed by the

U.S. Bureau of Standards and since practically all apparatus is nowlabeled in terms of milliliters, it is preferable to the older usage of the cubic


There are two reasons why any piece of volumetric apparatus is

marked with a temperature reading as well as with the number of milli-

liters which it delivers or contains. Water, and its solutions, vary in

density with change in temperature; thus a container which measures

100 ml. at 20 will not measure 100 ml. at other temperatures. Further-


Page 84: quimica inorganica cuantitativa


more, glass expands with rising temperature, and this too would alter the

capacity of the instrument. Almost all apparatus has a volume designatedfor 20. If used at any other temperature we must make allowance for the

change in volume which these two factors bring about.

Change of Volume of Solutions with Change of Temperature.The change in the volume of a solution due to a change in its temperatureis greater the more concentrated the solution. However, in quantitative

analysis one uses relatively dilute solutions. By the very definition of the

term density, the increase or decrease in volume will always be in the

ratio of the densities at the two temperatures. In the case of pure waterthe density at 20 is 0.99823 and at 25 is 0.99707. The difference, 0.00116

(in an average total of 0.99765 or approximately 1), is about 1.16 parts in

1000. Thus a quantity of water occupying a volume of exactly 1000 ml.

at 25 would have a volume of 1000.00 - 1.16 = 998.84 ml. at 20.A change of 5 in temperature would therefore introduce an error of morethan 1 part per 1000. In the case of solutions the error usually is greaterthan the above; solutions of most electrolytes of concentrations around0.1 M or 0.2M will so vary in volume that an error up to 3 parts per 1000

occurs for a 5 change in temperature. It is necessary therefore, as a

general rule, to maintain the temperature of our solutions within less

than 5 of that at which the apparatus was calibrated, and preferablywithin 2, if we wish to keep our precision within 1 part per 1000.

Change in Capacity of Class Apparatus with Change of Tem-perature. The cubic coefficient of expansion of glass is 0.000025 per 1.That is to say, for each change of 1 the glass apparatus will vary in

capacity by twenty-five millionths of its previous volume. This may be

expressed mathematically by the equation,

(1) V - V[l + 0.000025(<' - /)]

where V and V are the final and initial capacities of the glass container

at the final and initial temperatures /' and t respectively. For example, a

flask which has a capacity of 1000 ml. at 20 will contain at 25 a volumeof

V = 1000[1 + 0.000025(25 -20)]

= 1000.13 ml.

It is clear then that, since the above shows a variation of only slightlymore than 0.1 part per 1000, a change of 5 does not introduce a significant

error, except for the most precise work, as far as expansion or contractionof the glass apparatus is concerned.


Suppose that we wish to calibrate a liter flask. We may have either of

two ends in view. We may wish to learn the true volume which is con-

tained by the flask when filled up to the mark on its neck, or, on the other

Page 85: quimica inorganica cuantitativa


hand, we may wish to place a new etched line on the neck at that height

which actually does indicate exactly 1000 ml. In either case we must have

at hand a balance which is capable of weighing up to 2000 g.

First let us consider the determination of the true volume contained at

20 by a so-called 1000 ml. flask when filled to the graduation mark. Theflask is thoroughly cleaned with cleaning solution, rinsed first with tapwater and then three times with distilled water and dried. 1

Procedure. Place the stoppered flask on the left-hand pan of the

balance and add weights, or any metal pellets, copper shot for example,to the right-hand pan until an approximate balance is obtained, that is,

until the equilibrium point is close to the mid-point on the scale. Removethe flask and replace with weights on the left-hand pan until the same

equilibrium point is established. The sum of these weights represents the

weight of the flask, inasmuch as this substitution method of weighingcancels out any errors that otherwise would result from inequality in

length of the balance arms. Next carefully udd distilled air-free water to

the flask at room temperature, say 25. It is well to add the last milliliter

or two by means of a small pipet bringing the meniscus just tangent to the

graduation mark of the flask. Be careful that no droplets adhere to the

upper part of the flask. Replace the stopper and weigh flask and water,

again by the substitution method. From the weight of the water itself

the true volume of the flask is obtained from the density and temperature,corrections being made for buoyancy and glass expansion. The data of

Column 6, Table 4, are convenient for this purpose.

Next, let us say that we wish to calibrate the flask so as to contain at a

given temperature for instance, 20 a definite volume of liquid for

example, 1000 ml. After cleaning and drying the flask place it on the

left-hand pan and, along with it, weights the sum of which is equal in

grams to the number of rnilliliters of water the flask is to contain in the

present example, 1000 g. From an auxiliary set of weights, or with metal

pellets, add a tare to the right-hand pan and determine the equilibrium

point. Now remove all the weights, i.e., the 1000 g., from the left-hand

pan. Do not add water as yet to the flask for to do so would require 1000

g. of water and would ignore the effects of density change (the tempera-ture is not 20), the buoyancy effect and the effect of contraction of glass.

To compensate for these effects we must add weights to the left-hand pan.In the present case this amounts to 3.85 g., as may be seen from Table 4.

(The compensating weight is always added to the pan with the emptyflask since the figures in Column 6 are less than 1000.) After placing the

3.85 g. on the left-hand pan water is added to the flask until the original1 The drying may be accomplished quickly by rinsing the clean flask once or twice

with a small amount of alcohol followed by a rinse with ether. Then insert a glass tube

connected to a suction pump and draw cotton-filtered air into the flask to sweep outthe ether vapor.

Page 86: quimica inorganica cuantitativa


Table it




equilibrium point is established. The flask now contains that weight of

water at 25, namely, 996.15 g. which marks the height to which the

glass flask must be filled at 20 to contain 1000 ml. A mark is etched with

hydrofluoric acid at the level of the water meniscus.

Meaning of Table 4. The data of Table 4 are required to calibrate

glass apparatus in order to correct for the change in density of water with

change of temperature, for the buoyancy effect of air in weighing and for

the expansion or contraction of glass with change of temperature from 20.We have assumed, for the sake of illustration, that the temperature at

which we are calibrating a flask is 25. At this temperature the apparent

weight of 1000 ml. of water as taken from Column 6 of Table 4 is 996.15

g. Since the weights which were removed from the pan after obtaining the

original equilibrium point totaled 1000 g., the difference, or 3.85 g. (Col-

umn 5), represents our compensating weight and this must be added to the

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left-hand pan. Obviously if we now add water to the flask until we re-

establish the original equilibrium point we shall have added 996.15 g.

of water to the flask, and this is the amount of water which must be taken

at 25, our working temperature, to indicate in glass a volume of 1000 ml.

at 20. (Weights corrections, since they appear only in the fourth, or

sometimes the third, decimal place, need not be taken into account here

where the total weight involved is so large. Neither is it necessary to

ascertain equilibrium points better than approximately since an error of

one scale division or so would cause a corresponding weight error of only a

few milligrams, which in a total of about 1000 g., or even 100 g., is muchless than 1 part per 1000.)

To comprehend fully the meaning of the data of Table 4, let it be re-

stated that three considerations must be taken into account when wecalculate the proper weight of water to be used in volumetric calibrations.

These are : the density factor, the buoyancy factor and the glass expansionor contraction factor. We may now examine each of these.

Correction 1 (Column 2) is simply the density correction. Water at 4

has a density of unity and above this temperature, because of expansionof water, its density is less than unity. Thus above 4 the mass of water

which will occupy a volume of 1000 ml. will be less than 1000 g. ; the mass,

or in vacuo weight, will be lOOOrf, where d is the density at the higher tem-

perature. The figures of Column 1 are the product of 1000 ml. and the

density of water at the different temperatures. For example, the densityof water at 25 is 0.99707; this multiplied by 1000 gives the figure in

Column 1 for 25, namely 997.07 g. This differs from 1000 g., the weight of

a liter of water at 4, by 2.93 g. It follows that this value, 2.93 g., and the

others of Column 2, are the weights which must be subtracted from 1000

g. in order to obtain the weights of water in vacuo which at the indicated

temperatures occupy 1000 ml. If the density correction were the onlycorrection necessary, then for the calibration just outlined 997.07 g. of

water would have been taken.

The second correction is for the buoyancy effect of air upon the water

and the weights. Anything weighs less in air than in vacuo. The correction

to be made is the difference between the two buoyancy effects which

respectively are equal to the weights of air which are displaced by the

water and by the weights. The water displaces of course almost exactly a

liter of air, and a liter of air weighs approximately 1.19 g. (see p. 56,

Chapter 4), depending upon atmospheric pressure, temperature and

humidity. The volume of air displaced by the brass weights is equal to

their mass, 997 g. in our example, divided by the density of brass, 8.4

(if copper shot were used as a tare, 8.9, the density of copper). This gives

997/8.4 = 119 ml. This quantity of air weighs (119) (0.00119) = 0.14 g.

This weight subtracted from 1.19 g. gives 1.05 g. as the buoyancy correc-

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tion. (Incidentally, if this correction is added to Correction 1 and the sumsubtracted from 1000 g. we would have that weight of water to be taken

for calibrating at 25 a flask which subsequently was to be used at 25.)These corrections, for buoyancy, are found in Column 3 of the table.

The third correction takes care of the expansion or contraction of the

glass container (expansion or contraction, depending upon whether the

calibration is carried out above or below 20, the temperature taken bythe Bureau of Standards as the standard temperature for calibrations).

It already has been pointed out in equation (1) just how the capacity of a

glass container changes as the temperature is varied. We saw that this

does not introduce an error greater than 1 part per 1000 when the tem-

perature differs by 5 or less from standard temperature. However, in

the calibration of apparatus we should aim for the greatest possible pre-

cision. Correction 3 thus, while small (compare Columns 2, 3 and 4 of

Table 4), should be included. It was shown previously (p. 66) that a glass

vessel holding 1000 ml. at one temperature would contain 1000.13 ml. at

a temperature 5 higher, an increase of 0.13 ml. Now 0.13 ml. of water

having a density of 0.99707 weighs 0.13 g. This is Correction 3, which ob-

viously must be subtracted for a working temperature which is 5 above

standard, 20. (Likewise the same figure, 0.13, is added for a working

temperature which is 5 below standard; see Correction 3 of Table 4

for 15.)

We may avoid the laborious process of making these three calculations

by using Table 4 where the several corrections have been assembled for

usually prevailing temperatures. From Column 6 one may obtain the

weight of water which must be weighed in air with brass weights2 at any

temperature shown in the table in order to indicate a volume of 1000 ml.

in a glass container at 20. Column 5, giving the sum of all three correc-

tions, tells us the number of grams to add to the left-hand pan as "com-

pensating weights" previously mentioned. If the flask to be calibrated is

not to contain 1000 ml., then a proportionate part of the numbers of

Columns 5 and 6 apply; for example, to calibrate at 25 a flask to contain

500 ml. at 20, the figure for Column 5 would be half of 3.85 or 1.93 g.


Just as in the case of calibrating a flask, we may wish either to

determine the true volume of the pipet as designated by its graduationmark, or we may wish to place upon the pipet a new mark so that a

certain quantity of liquid will be delivered.

Procedure, hi the first case we proceed as follows: Aftor the pipet is

cleaned so that it shows a uniform film after rinsing, fill by suction with

2 A difference of only 0.01 part per 1000 results if copper shot are used as a tare

instead of auxiliary brass weights; prove this.

Page 89: quimica inorganica cuantitativa


distilled water at room temperature so that the meniscus is above the

graduation mark. Quickly place the forefinger over the top in order to

retain the liquid in the pipet. Hold the pipet in a vertical position over a

beaker and carefully admit air by releasing the pressure of the finger,

thus allowing the level of the water to fall until the meniscus is tangentto the mark. Remove any hanging droplet by touching the tip to the

side of the beaker or by wiping with a piece of Kleenex tissue. Next runthe water from the pipet into a clean, dried weighing bottle which hasbeen weighed by the substitution method to the third decimal place.Allow the pipet to drain for about 15 seconds, then touch the tip to the

inside surface below the ground-glass neck to remove the hanging drop.Do not blow or tap out the small amount of water remaining in the

pipet tip. Weigh the bottle and water, again by the substitution method,and from the data of Table 4 calculate the true volume.

In the second case, to calibrate the pipet for the delivery of a pre-determined volume, the procedure is somewhat different. Determine the

true volume of the pipet exactly as described in the above paragraph.If this proves to be greater than the volume specified by the manu-facturer, place a gummed label on the stem at a position somewhatbelow the etched mark; if the true volume is less than the nominal value,

place the label somewhat above the etched mark. Determine the true

volume of water delivered from this new marker. We now have two true

volumes; one represents the volume delivered from the manufacturer's

mark and the other the volume delivered from the gummed label.

From the distance between these two marks and the true volumes corre-

sponding to each, estimate by interpolation the position of a third mark,between the two, which should give the desired volume. Paste a newlabel at this point and make a new determination of the true volumedelivered by the pipet from this level. Sometimes due to slight errors in

interpolation it may be necessary to shift the final label slightly in order

to secure the proper position for the desired definite volume. When the

position has been accurately established substitute a permanent etched

mark for the gummed label.


A buret may show deviations from the true volume at any position

throughout its length. It therefore is necessary to check the buret at

several points. For a 50 ml. buret it is usual to determine the true volumewhich is delivered at 10 ml. intervals.

Procedure. After the buret has been thoroughly cleaned, place dis-

tilled water at room temperature in the buret and lower the meniscusuntil it is tangent to the zero mark. See that no air bubble is in the

buret tip and remove any hanging droplet. Then open the stopcock

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slightly so that the water flows drop by drop into the weighing bottle

which previously has been cleaned, dried, cooled and weighed with its

stopper to the third decimal place (substitution method of weighing).

When the meniscus has fallen exactly to the 10 ml. mark turn off the

stopcock. Handling the weighing bottle with tongs, detach any hanging

droplet by touching the buret tip to the inside surface of the weighingbottle below the ground-glass neck. Replace the stopper and weigh the

bottle plus water to the nearest thousandth gram. The true volume

delivered between the zero and 10 ml. mark is obtained, as usual, by

dividing the weight of water by Kooo of the figure found in Column 6

of Table 4 at the proper temperature. The difference between the volume

so calculated and the buret reading, namely 10.00 ml., is the buret


Refill the buret to the zero mark. The above procedure is now repeated

except that water is withdrawn until the 20 ml. mark is reached. Then,after refilling the buret, 30 ml. is withdrawn and weighed; then 40 ml.,

and finally 50 rnl. In every case the interval being calibrated should be

run in duplicate and duplicate runs should check within 0.01 ml. After

the data for all five intervals have been obtained construct a graph in

which buret readings are plotted against the corrections. A sample set of

data for the calibration of a buret are given in the following table.

Table 5


In any calibration of volumetric apparatus all weighing bottles must,of course, be handled with tongs. In making duplicate determinations

time can be saved by always having one weighing bottle in the oven at

105 for drying and then in the desiccator for cooling, during the time

that the other weighing bottle is being weighed. Note that weighingsare made only to the third decimal place. Burets can be read only to the

hundredth milliliter and if, during a calibration, we determine the weightof water to the thousandth gram (and the corresponding volume to the

Page 91: quimica inorganica cuantitativa


thousandth milliliter) and then round off the volume to the hundredth

place we drop the last, doubtful figure and thus have the true volume,

theoretically at least, with no doubtful figure.

Problem. A flask filled to the mark at 24 contains 996.10 g. of water whenweighed in air with brass weights, (a) What is its volume at 24? (b) What is its

volume at 20?Solution, (a) From Table 4, the mass of water which in air at 24 weighs 996. 10 g.

is 996.10 + 1.05 = 997.15 g. At 24 the mass of 1 1. of water is 997.32. ThereforeQ97 15

the volume of the flask at 24 is' = 999.83 ml.

\)y < .<J4

Or, part (a) may be solved directly from Table 4 as follows: Because the

capacity is wanted for the same temperature as that at which the calibration wascarried out, Correction 3 is not included. The volume of the flask at 24 is 996.10

+ Correction 1 -f Correction 2. The sum of these corrections at 24 is 3.73. Thusthe volume at 24 is 996.10 + 3.73 = 999.83 ml.

(b) If the flask contains 999.83 ml. at 24 we see by Column 4 of Table 4 thatit will contract in capacity to an extent of 0.10 ml. when the temperature is 20.Therefore the flask will have a volume of 999.73 ml. at 20.

Or, part (b) may be solved directly from Table 4 in the following manner. Thesum of all three corrections, Column 5 of Table 4, at 24 is 3.63. Therefore thevolume of the flask at 20 will be 996.10 + 3.63 - 999.73 ml.


1. What weight of water must be used at 27 to locate the mark on a flask so thatit will serve as a calibrated flask of 1000 ml. capacity at 20?

Answer: 995.67 g.

2. What weight of water must be used at 30 to locate the mark on a flask so thatit will serve as a calibrated flask of 500 ml. capacity at 20?

Answer: 497.44 g.

3. A flask is filled to the mark by adding 498.50 g. of water at 22. What is the

capacity of the flask if used at 22?Answer: 500.13 ml.

4. A flask which is supposed to be of 1 1. capacity is filled to the mark with waterat 18 weighing 997.51 g. (a) What is its volume at 18? (b) What is its volumeat 20?Answer: (a) 999.95 ml; (b) 1000.00 ml.

5. A solution is standardized by titration in a glass buret and found to be 0.1015N. The temperature is 20. Later it is restandardized but the temperature is

26. What normality should be obtained in the latter determination?Answer: 0.1014 N if the standardization was primary.

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Chapter 6


by volumetric methods are based upon the interaction of

the constituent to be determined with a definite volume of a solution

of known concentration, called a standard solution. The determination

of the concentration of the latter solution is known as the standardiza-

tion of the solution. The addition of an equivalent quantity of one solu-

tion to a definite amount of another and the measurement of the volume

employed is called a tiiration. The point at which an equivalent quantityof the one solution has been added from the buret is recognized by the

use of an indicator. This is a substance which, present in small amountin the solution being titrated, will reveal, usually by a change in color,

the point at which enough of the titrating solution has been added to

complete the desired chemical reaction. Indicators thus mark the end

point of a titratioii. Ideally the end point should coincide with the

equivalence point, that is, the stoichiometric point, at which chemically

equivalent quantities of the reacting substances have been brought

together. This is the object of all titrations, and a suitable indicator

for a given titration is one which will give the stop signal when chemi-

cally equivalent quantities have reacted. As will be shown later, the end

point and the stoichiometric point do not always coincide exactly but

for a feasible titration the two must closely approach one another.

Methods of Expressing Concentration. The concentration of a

solution may be expressed in several ways, all of which reveal the

quantity of solute per unit volume or per unit weight of solution or of

solvent. Obviously one method is to state the weight of solute per liter

or milliliter of solution. For example, we might have a solution of sodium

chloride containing 50 g. of the salt in a liter of aqueous solution. Againthe concentration may be stated in terms of weight of solute per unit

weight of solution, as 50 g. of sodium chloride per 1000 g. of solution.

Such a solution would be a 5 per cent solution. This would be not quitethe same as the former since the volume of a given amount of solvent

will change somewhat upon dissolving the solute. Another method

essentially the same as this is to give the specific gravity and the ^weight

percentage of the solute present. The statement that a sulfuric acid


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solution has a specific gravity of 1.84 and consists of 96 per cent H 2S0 4

by weight is unambiguous since there is only one concentration of

sulfuric acid compatible with these data. Another common expression

of concentration is per cent by volume. A solution of 10 ml. of alcohol

per 100 ml. of solution is said to be 10 per cent alcohol by volume.

Although all of the above means of denoting the concentration are

frequently employed, none of them meets the particular needs of volu-

metric analysis so well as a method based upon the interacting chemical

units instead of upon absolute weight or volume percentages. If it were

true that 1 g. of a substance always reacted with, say, I g., or 2 g.,

of another substance, then it would be advantageous to make up all of

our standard solutions with a given number of grams of solute per unit

volume of solution. The fact is, however, that substances react with one

another in the ratio of whole numbers of molecules.

.Molar Solutions. If standard solutions were made up so as to con-

tain equal numbers of molecules per litei; of solution it follows that all

such solutions, if they react with one another, would do so in the ratio

of small, whole number volumes. This would make matters simpler than

if all the solutions contained the same number of grams per unit volume

of solution. Since all gram molecular weights contain the same number

of molecules' wejmay prepare such molar solutions by dissolving per

liter of solution the molecular weight of the solute expressed in grams.

That is to say, a molar solution is one which contains 1 gram molecular

weight, or I mole, of solute in 1 Li of solution. For example, 58.45 g. of

sodium chloride, 98.08 g. of hydrogen sulfate and 98.00 g. of hydrogen

phosphate all give, if dissolved in a liter of solution, molar solutions.

However, even the molarity system is not always the most convenient

means of expressing the concentrations in volumetric analysis. It would

be if it were true that chemical reactions always involved equal numbers

of molecules. But since, for example, hydrochloric acid and sodium

hydroxide react molecule for molecule, while sulfuric acid and sodium

hydroxide react, for complete neutralization, in the ratio of one molecule

to two, it would be better if we devise a system in which two solutions

having concentrations of the same numerical value will react volume for

volume. This may be accomplished if we employ the normality system

for concentrations.

Normal Solutions. It has been stated that if two substances react

at all they do so by whole numbers of molecules. But what is it which

determines whether the ratio in which molecules react will be one to one,

or one to two, or something else? In the case of an acid reacting with a

base the answer lies in the number of hydrogen arid hydroxyl ions per

molecule. An acid like sulfuric which ionizes to yield two hydrogen ions

obviously may react with two molecules of sodium hydroxide which

Page 94: quimica inorganica cuantitativa


dissociates to give only one hydroxyl ion since the essential reaction of

neutralization is

H+ + OH- - H 2

It follows that 1 1. of a solution of sulfuric acid containing half as manysolute molecules as a liter of sodium hydroxide solution would justneutralize the latter because the sulfuric acid would yield the samenumber of hydrogen ions as the base would furnish of hydroxyl ions.

Thus a 0.5 molar solution of sulfuric acid is exactly equivalent to a 1

molar solution of sodium hydroxide. Since they are equivalent there

must be a means of designating such equivalence. This is accomplishedby saying that the above are both 1 normal solutions. A normal solution

is defined as a solution 1 1. of which contains, or will replace or combine

with, 1 gram atomic weight of hydrogen (1.008 g.) or its equivalent. Sucha quantity of solute is called its gram equivalent weigh^. That is to say,

the gram equivalent weight of any acid, base or salt is the weight of the

substance in grams which will react with 1.008 g. of hydrogen or hydro-

gen ion, with 35.46 g. of chlorine or chloride ion, etc., and the normalsolution is one containing 1 gram equivalent weight of solute per liter of


It is easy to compute the equivalent weight of electrolytes. A few

examples will suffice to illustrate. Hydrochloric acid has a molecular

weight of 36.46 to which hydrogen contributes 1.008; therefore the useof 36.46 g. of hydrogen chloride will furnish 1.008 g. of hydrogen, andthe gram molecular weight and the gram equivalent weight are identical,

namely, 36.46 g. Sulfuric acid, molecular weight 98.08, will yield two

hydrogen ions per molecule; therefore 98.08 g. furnish 2.016 g. of hydro-gen ion or 49.04 g. furnish 1.008 g. The gram equivalent weight of .mlfuric

acid is thus 49.04 g. The gram equivalent weight of bases may be calcu-

lated on the basis of the number of hydroxyl ions from each moleculesince one hydroxyl ion is equivalent to one hydrogen ion. For sodium

hydroxide, molecular weight 40.01, the gram equivalent weight is 40.01 g.;for calcium hydroxide, molecular weight 74.10, the equivalent weight is

37,05. In the case of salts one may judge matters by the number andvalence of either of the ions which the salt yields. If it dissociates to giveone univalent ion of a given sign the molecular weight and the equivalentweight are the same, as in the case of sodium chloride, 58.45. But sodiumsulfate, molecular weight 142.1, yields two univalent positive ions, whichare equivalent to two hydrogen ions, and its gram equivalent weight is

therefore 71.05 .g. Aluminum sulfate, molecular weight 342.1, would bythe same reasoning have an equivalent weight equal to one-sixth of its

molecular weight, or 57.02.

The terms milliequivalent weight, meg., and gram milliequivalent

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weight are often used. As the term implies, the meq. of a substance is its

equivalent weight divided by 1000. Thus the meq. of NaOH is 0.04001

and its gram meq. is 0.04001 g.

Objection is sometimes raised to the normality system of designatingconcentration because the equivalent weight of a substance may varyaccording to the particular reaction which it undergoes. For example,the equivalent weight and the molecular weight of sodium carbonate are

the same if the reaction proceeds according to the equation

Na2C0 3 + HC1 = NaHC 3 + NaCl

but the equivalent weight is only half the molecular weight if conditions

are such that the reaction is

Na 2C0 3 + 2HC1 = H2C0 3 + 2NaCl

Furthermore, the equivalent weight of potassium permanganate,which as a salt evidently is identical with its molecular weight, is, as

we shall see later, a fifth of the molecular weight when this substance

acts as an oxidizing agent in an acidic environment, and is a third of the

molecular weight when it functions as an oxidant in a basic solution.

However, the variability of the value for the equivalent weight of manysubstances, instead of being a source of confusion, is actually a means bywhich one comes to a quick understanding of the fundamental principlesof equivalence. It is only necessary to consider the particular chemical

reaction which the reagent is undergoing and to ascertain therefrom the

number of hydrogen equivalents involved in the reaction. This number,divided into the molecular weight of the acid or base or salt, yields the

equivalent weight. The advantage in volumetric analysis of the volumefor volume relation when the normality system is used far outweighsthe simplicity which is sometimes claimed for the more primitive mo-

larity system ; furthermore, the latter does not remove all ambiguity. In

many cases it would be open to doubt as to what is meant by a molarsolution unless some explanatory note we ^ given. For example, a moleof water is not 18 g. unless the formula of water is H 2 and not (H20) x .

Mso the mole of a salt which forms one or more hydrates obviously maybe any of several values.


A standard solution is one of definite and known concentration. Theconcentration is usually designated in terms of the normality of the

solution. If the solution contains 1 gram equivalent weight of the solute

per liter of solution it is normal, 1 AT, A solution which contains one-

tenth of a gram equivalent weight of solute per liter is 0.1 N. Thus 40 g.

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of sodium hydroxide in a liter of solution makes a 1 TV solution, 4 g. perliter gives a solution which is 0.1 TV, and 0.2 g. per liter makes the solu-

tion 0.005 AT, etc. A standard solution often may be prepared in either

of two ways. The proper amount of solute may be weighed exactly and

dissolved in a carefully measured volume of solution (by using a vol-

umetric flask). Or the solution may be made up using the approximatelycorrect quantity of solute and solution and the exact value of the not;;

mality then determined by a process called standardization.

To illustrate the first method, assume that a tenth-normal solution

of sodium chloride is wanted. It may be prepared by weighing 58.45/10,or 5.845 g. of pure dry sodium chloride, dissolving this in pure water,

transferring quantitatively to a 1 1. volumetric flask, diluting to the

mark and shaking. This involves the somewhat troublesome task of

weighing the exact quantity, 5.845 g. This method cannot be employedif the solute is hygroscopic or if it cannot be obtained in a pure state.

In the second, more commonly employed method, an amount of the

solute close to the calculated quantity is weighed on a trip balance for

example, about 4 g. of sodium hydroxide and dissolved in roughly 1 I.

of solution. The solution so prepared will not be 0.1000 TV but nearly so.

Usually it is not important that the standard solution be exactly

0.1000 N, 0.01000 AT, etc., but only that the solution be close to the

desired normality and that its exact normality be known. The exact

normality of the solution is determined by titrating either against a suit-

able quantity of a "standard" (primary standardization) or against a

solution the exact normality of which is already known (secondary

standardization). Tf a solution of a certain definite normality for

example, exactly 0.1000 N is desired, a slightly more concentrated

solution is made up and standardized. Next it is diluted with water to

the extent calculated to establish the exact desired normality and then

restandardized (see p. 83).

Primary Standardization. A standard is a substance which will re-

act quantitatively with the solution to be standardized in such a mannerthat the result of the titration may be used to calculate the normality of

that solution. A standard must meet certain criteria.

1. The compound must be above 99.9 per cent pure, or of definitely

known composition. Usually it is a solid.

2. It should be a stable compound which maintains a constant com-

position in contact with air.

3. It should undergo drying at 105 without change in its composition.4. It must be capable of accurate weighing. For example, it must not

be hygroscopic and should not be easily sublimated, nor should its

physical state be too fluffy.

5. It should dissolve readily in water or in other suitable solvents.

Page 97: quimica inorganica cuantitativa


6. It must react with the solution to be standardized quantitatively,

according to one invariable reaction.

7. It should have a fairly large equivalent weight. Since almost all

student burets are of 50 ml. capacity a titration should not exceed this

volume of solution and usually will run between 30 and 40 rnl. This means

that there is a definite upper limit for the number of gram milliequivalent

weights of standard that may be used. If the rneq. of the standard is

small the actual weight appropriate for the titration of an approximately0.1 N solution may be less than 200 mg., which means, when it is remem-

bered that the error in weighing a substance may amount to 0.2 mg., that

a relative error of more than 1 part per 1000 would be introduced.

After a standard meeting the above requirements has been chosen, the

quantity necessary for a titration of 30 to 40 ml. of solution must be

calculated. A.S is true in so many cases, there are long methods and a short

way to make the calculation. It is always true that the amount of stand-

ard needed is equal to the product of three factors, namely: the approxi-

mate volume in liters (of the unknown solution) suitable for the titration,

the approximate normality of that solution and the gram equivalent

weight of the standard. No shorter method of calculation is possible. Toillustrate: if a solution of sodium hydroxide which is about 0.1 N is to

be standardized by titrating it against benzoic acid as the standard, the

weight of benzoic acid necessary for a titration involving around 30

ml. of the base is

(3 ^ooo)(0.1)(122) = 0.366 g.

Similarly, if sodium carbonate is to be used to standardize a 0.2 TV solu-

tion of hydrochloric acid one should weigh out about

- (3 %ooo)(0.2)(io^) = 0.318 g.

The above type of calculation illustrates a perfectly general method for

the reason that, since 1 1. of 1 TV solution contains 1 gram equivalent

weight of solute, 30 ml. of approximately 0.1 N solution contains about3?fooo of 0.1 of 1 gram equivalent weight. Therefore about 3%ooo of 0.1

of a gram equivalent weight of the standard will be required for the

reaction taking place during the titration.

It must be emphasized that although only a rough calculation should

be made to ascertain the quantity of standard suitable for the titration

and that the amount actually weighed out need only be somewhere near

this calculated quantity, the weighing must be made exactly and to the

fourth decimal place. The sole object of the standardization is to learn

the normality of the solution with a possible error, as a rule, of not over

1 part per 1000. Therefore the standard must be weighed to the fourth

decimal place and the buret must be read to the hundredth milliliter.

Page 98: quimica inorganica cuantitativa


One should avoid discarding a weighed portion of standard if it happensto be somewhat more than the calculated amount for example, 0.366 g.

in the case of the benzoic acid-sodium hydroxide titration cited above

for it is obvious that if the base is close to 0.1 TV it would require

(5%ooo)(0.1)(122) = 0.61g.

of benzoic acid to render refilling of the buret very likely.

The standard may be weighed, of course, either by difference or by

using balanced watch glasses (see p. 63). If the titration is to be carried

out in an Erlenrncyer flask the former is preferable because of the ease of

transferring the standard from the weighing bottle or vial to the flask. If a

beaker is used for the titration it is better to use balanced watch glasses

for several reasons. First, it is a simpler matter to weigh out a quantity of

standard between the upper and lower limits of the amount wanted, i.e.,

a quantity which will require between 30 and 50 ml. of solution from the

buret. Secondly, the quantitative transfer of the standard to the titrating

beaker is accomplished without loss simply by lifting with tongs the watch

glass containing the accurately weighed standard and carefully placing it

within the titrating beaker. The solvent then is added and the titration

completed without removing the watch glass. Another advantage of this

procedure is that stirring may be effected during the titration through

rocking the watch glass by touching f>st one side and then the other with

a glass rod. This diminishes the amount of carbon dioxide introduced

from the air, an occurrence which happens if the solution is too vigorously

agitated with a stirring rod. The absorption of carbon dioxide is objection-

able in acid-base titrations when an indicator such as phenolphthalein is

used (see p, 125). Sometimes an analyst uses a camel's-hair brush to

sweep the material from the watch glass to the titrating vessel, but, while

this procedure is safe in experienced hands, it involves the possibility of

loss unless utmost caution is taken to dislodge all particles from the brush.

Calculation of Exact Normality. When the proper amount of the

standard has been weighed and transferred to the titrating flask and the

titration completed, the volume of tho solution used is read, to the hun-

dredth milliliter, from the buret. The calculation of the normality is quite

simple and one should avoid becoming involved in unnecessarily long

computations. An example will illustrate the point. Suppose a solution of

sodium hydroxide approximately 0.1 TV has been standardized againstbenzoic acid weighing 0.4134 g., and that the end point for the titration

was 33.52 ml. From these data it is possible to calculate the normality of

the base by carrying out the following logical but unnecessary number of



1 Some psychologists claim that it is poor pedagogy to illustrate a wrong, or an un-

desirable, method. However, the involved calculation of normality here illustrated is

Page 99: quimica inorganica cuantitativa


1. Compute the grains of benzole acid equivalent to 1 1. of sodium

hydroxide.2. Compute the grams of sodium hydroxide equivalent to the grams

of benzoic acid of step 1.

3. Divide the weight of sodium hydroxide of step 2 by the equivalent

weight of sodium hydroxide. This will give the number of gram equivalent

weights of sodium hydroxide per liter of sodium hydroxide solution andtherefore is the normality of the solution.

In the case at hand these calculations are, in order:

Step 1. ~~(1000) = 12.33 g. C 6H5COOH

Step 2. ~J (12.33) = 4.041 g. NaOH

Step 3. ~ = 0.1010 = Normality of NaOH

The above computation is more quickly made if one realizes that the

normality of the titrated solution always must be

,..,,, -

,. . , g. of standard =c= 1 1. of unknown solution

(1) TV of solution titrated =--. , 7-r-rr --

, i---

equivalent weight of standard

From the data used in the previous illustration this yields:

0-4134QO co

N of NaOH solution =

Obviously this shorter calculation merely eliminates the multiplication by40.01 in step 2 and the subsequent division by 40.01 in step 3 of the longermethod.

Equation (1) must yield the normality. There should be no confusion

over the fact that we divide one set of numbers pertaining to the standard

(here, benzoic acid) by another number also pertaining to the standard

and arrive at a quotient (the normality) pertaining to the unknown solu-

tion (here, sodium hydroxide). If one contends that this quotient, 0.1010,

represents the normality of a hypothetical benzoic acid solution it need

only be borne in mind that such a benzoic acid solution would be one of

which a liter was exactly equivalent to a liter of the sodium hydroxidesolution. Therefore it would, of necessity, have the same normality as the

sodium hydroxide solution. Equation (1) thus is valid for the calculation

so common among students of elementary quantitative analysis that it is believedbest to contrast this method with the shorter, recommended method which follows

(Equation (1)).

Page 100: quimica inorganica cuantitativa


of the normality of the unknown solution and it will be found, in the

chapter on oxidation and reduction, that this equation is applicable to

the calculation of the normality of those solutions as well as to solutions

of acids and bases and salts. It should be mentioned that primary stand-

ardizations are usually carried out in triplicate.

Secondary Standardizations. It has been mentioned already that

the normality of a solution may be determined by means of a secondary

standardization. That is to say, if a standard solution of hydrochloric

acid, for example, is available, the normality of a solution of a base maybe determined by titration of the base against the acid. In any titration

the product of the normality and the volume of the one solution must

equal the product of normality and volume of the other. To illustrate, it is

obvious that 100 ml. of 0.1 N base must be equivalent to 10 ml. of 1 Nacid, so that (100)(0.1) = (10)(1), or, in general,

(2) NV = N'V

Thus if 30.00 nil. of hydrochloric acid solution known to have a normalityof 0.1089 were titrated with the sodium hydroxide solution already con-

sidered and the end point occurred at 32.35 ml. of NaOH, the normalityof the base would be calculated as follows:

Base Acid

N . V = AT/. V

JV(32.35) = (0.1089) (30.00)

N = 0.1010

This type of standardization is called secondary because the normalityof the unknown is determined by titrating against a solution which itself

was standardized against a primary standard. Secondary standardiza-

tions are subject to two sources of error: in addition to any error made in

the actual titration of the unknown solution, those errors acquired duringthe primary standardization of the hydrochloric acid, in the case at hand,will be carried along and included in the normality of the unknown, here

the sodium hydroxide. Accordingly, it is customary, if two good checks

are obtained in secondary standardizations, to make the determination

only in duplicate. Since the errors, determinate and indeterminate, in

every standardization may become additive, secondary standardizations

should be avoided when highest accuracy is desired.

It should be emphasized that in the equation, NV = N' V, the

product of the normality and the volume, in milliliters, gives the numberof gram milliequivalent weights of the substance. We know, for example,that 1000 ml. of 1 N solution contains 1 gram equivalent weight of solute,

which is to say 1000 gram milliequivalent weights; and the product,

NV, is (1)(1000) = 1000. Similarly, 50 ml. of 0.2 N solution gives the

Page 101: quimica inorganica cuantitativa


NV product: (0.2) (50) = 10 or 10 gram milliequivalent weights.2 Re-

membering this, such problems as the following become simple to solve.

Problem. A sample of oxalic acid, H 2C 2O4, weighing 0.2268 g. is titrated against0.2000 N NaOH solution, 25.14 ml. being required. Calculate the purity of the

oxalic acid.

Solution. (0.2000) (25.14) = 5.028 gram meq. NaOH used. Thus there must be

5.028 gram meq. H 2C 2O4 in the sample. The gram meq. of H 2C 204 is 0.09004/2 g.

or 0.04502 g. Therefore

(5.028) (0.04502) = 0.2264 g. H 2C 2 4


(T2268(100) =99.8% pure

Of course the last two of the above three steps may he combined.

Another way of looking at the above problem is from the standpoint of equa-tion (1) from which we have by substituting in the equation

0.2000 =25.14

where X is the purity of the oxalic acid. Solving, we obtain

X = 0.998 or 99.8%

Dilution of Solutions to a Definite Normality. Sometimes it is

desired to prepare a standard solution of a definite normality. Often it is

impossible to weigh exactly the proper amount of solute as, for example,in the preparation of sodium hydroxide solution where the solute is hygro-

scopic. Tn any event the weighing of a definite quantity of a substance to

the tenth milligram is troublesome; it usually is simpler to prepare a solu-

tion of slightly greater concentration than desired and then to dilute it to

the proper concentration. To illustrate, if an exactly 0.1000 N solution of

sodium hydroxide is to be prepared, the solution mentioned in the pre-

ceding section having a normality of 0.1010 may be utilized. Again equa-

tion (2) will serve for calculating the dilution necessary. If we wish 1 1. of

0.1000 N solution the question is simply this: How much of the 0.1010 Nsolution must be taken and diluted to 1000 ml. in order that the resulting

solution will be exactly 0.1000 TV?

AT . V = N' . V0.1010V = (0.1000) (1000)

V = 990.1 ml.

Therefore if 990.1 ml. of the 0.1010 N solution is diluted to 1000 ml. the

resulting solution will be exactly 0.1000 N. After making such dilutions

the new solution should be standardized in order to confirm the expected


2 The terms milliequivalent, milliequivalent, weight and grain milliequivalent

weight are often used interchangeably.

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The analyst often finds it necessary to prepare solutions of common

reagents like sulfuric, nitric, hydrochloric and acetic acids, arid am-

monium hydroxide from the concentrated solutions purchased from the

manufacturer. The bottles containing these concentrated solutions are

labeled with the specific gravity and the percentage by weight of the con-

stituent. From these data the normality of the concentrated solution

may be roughly calculated and by equation (2) the volume to be taken to

prepare a more dilute solution may be computed. Table 6 gives the data

for the common acids and for ammonium hydroxide. /

Table 6



IF 2S() 4

IIC1 .

1IN0 3 .

I1C 2 IT,<)


The figures of the third and fourth columns of the table may be calcu-

lated from those of the first two columns. This may be illustrated with the

case of H 2S0 4 . The specific gravity is 1.84, which means that a rnilliliter

of the concentrated acid weighs 1.84 times as much as a milliliter of water,

or 1.84 g. Since 96 per cent or 0.96 of the concentrated solution is hydro-

gen sulfate, the remainder being water, (0.96) (1.84) = 1.77 g. of EUSO*is in 1 ml. of the solution. A liter of the solution then contains 1770 g.

of H 2SO 4 . This divided by the gram equivalent weight, or 177%9 = 36.1,

gives the normality of the concentrated acid.

If one wished to prepare 1000 ml. of a 0.5 TV solution the volume of

concentrated acid necessary obviously would be

AT . v = N' . V36.1F= (0.5) (1000)

V = 13.9 ml.

That is, if 13.9 ml. of the concentrated sulfuric acid is diluted to 1000 ml.

the resulting solution would be 0.5 TV. The exact normality, to four sig-

nificant figures, then could be determined by standardization.


1. How many grams of each of the following suhstances are necessary to preparethe solutions indicated, assuming in each case that the acid resulting after

dissolving is to be completely neutralized? (a) 1 1. of 0.5000 N

Page 103: quimica inorganica cuantitativa


(b) 500 ml. of 0.1000 TV H 2S0 4 . (c) 100 ml. of 0.2000 TV acid made from dis-

solving P 2 5 in water, (d) 1 1. of 0.5000 N NaHS0 4 .

Answer: (a) 30.03. (b) 2.452. (c) 0.4732. (d) 60.03.

2. One mole of the alum, A1 2(S0 4)3(NH 4 ) 2SO4.24H 2 contains how many (a)

moles of water; (b) grains of water; (c) gram atomic weights of aluminum;(d) gram molecular weights of NH 3 ; (e) grams of oxygen?Answer: (a) 24; (b) 432.38; (c) 2; (d) 2; (e) 640.

3. An aqueous solution of hydrochloric acid has a specific gravity of 1.120 and is

23.80 per cent HC1 by weight. Calculate the normality of the solution.

Answer: 7.31 N.

4. Two solutions of sulfuric acid are available. One is 0.2000 N and the other is

0.2500 N. How many milliliters of the two should be brought together in

order to prepare 1000 ml. of a solution about 0.2222 N? (Hint: use the equa-tion, NV + N'V = N"V".)

Answer: 556 ml. of 0.2000 N plus 444 ml. of 0.2500 N.

5. What must be the normality of a potassium hydroxide solution 100 ml. of

which, added to 250 ml. of a 0.1500 N KOH solution, will give a resultingsolution which is 0.2000 TV?

Answer: 0.3250 N.

6. A sample of 0.1928 g. of pure, dry oxalic acid is used to standardize a sodium

hydroxide solution. If 32.82 ml. of the base is necessary in the titration whatis the normality of the sodium hydroxide?

Answer: 0.1305 N.

7. How many milliliters of the sodium hydroxide solution of problem 6 would beneeded to titrate a sample of 0.6060 g. of pure, dry potassium acid phthalate,KIIC 8H 4 4 ?

Answer: 22.74 ml.

8. What is the normality of a solution of potassium hydroxide of which 29.16

ml. is equivalent to 40.08 ml. of a potassium tetroxalate solution made by dis-

solving 4.237 g. of the KHC 2O4.H 2C 2()4.2II 2O in water and diluting to 500


Answer: 0.1371 N.

9. Calculate the equivalent weight of an acid of which 0.9250 g. is equivalent to

40.00 ml. of a sodium hydroxide solution if 5.00 ml. of the base is equivalentto 0.05025 g. of Na 2CO 3 .

Answer: 122.L

10. A sample of 0.3223 g. of calcite, CaC0 3 , is dissolved by gently heating in

100.0 ml. of 0.1000 N HC1. After the reaction has ceased the excess acid is

titrated with 0.2000 N NaOH, 18.51 ml. being required. Calculate the per cent

purity of the calcite.

Answer: 97.8%.

11. A mixture of pure sodium carbonate and pure calcium carbonate weighs0.3000 g. A volume of 40.80 ml. of 0.1401 N acid is necessary to neutralize

exactly the mixed carbonates. Calculate the per cent of each present in the


Answer: 84% Na 2C0 3 ; 16% CaC0 3 .

Though four significant figures are given in the data of this problem the an-

swers are limited to two significant figures. Why?

Page 104: quimica inorganica cuantitativa

Chapter 7


X HE REACTION between an acid and a base essentially consists of the

combination of hydrogen ions with hydroxyl ions to form water. Such

reactions are known as neutralization reactions. According to classical

theories an acid is a substance which, in solution, furnishes hydrogen ions,

and a base is one which furnishes hydroxyl ions. A strong acid or base is

one which is practically completely dissociated in aqueous solution, while

a weak acid or base is one which is only slightly ionized.

Bronsted 1 extended the concept of acids and bases in order to take

into account their variations in strength when dissolved in different

solvents and the part played by a particular solvent in ionization. Accord-

ing to the Bronsted concept an acid is defined as a substance which fur-

nishes a proton (a positively charged hydrogen atom), that is, gives up a

proton or is a proton donor. A base is defined as a substance which accepts

a proton, that is, combines with a proton or is a proton acceptor. These

definitions of an acid and a base may be summed up in the equation,

HA * H+ + A-acid proton base

If the above reaction is followed by a combination of H4* and the

solvent, then the solvent is functioning as a base.

HA + H 2O + H30+ + A-acid base acid base

Here water acts as a base since water accepts a proton to form the hydron-ium ion, H3O+ (also called the oxonium ion). From the following equa-tions for reactions which may take place in aqueous solution it is evident

that a given substance may act sometimes as an acid and at other times

as a base.

(a) H 2 =* H+ + OH-(b) H 30+ *= H+ + H 2O(c) H 2CO 3 ;F H+ + HCOr(d) HCOr ^ H+ + CO a


1 Bronsted, Rec. Trav. Chirn., 42, 718 (1923); J. Phys. Chem., 30, 777 (1926); Chem.

Revs., 5, 231 (1928); Trans. Faraday Soc., 25, 59 (1929); Lowry, Chem. & Ind., 42,43 (1923); Hall, J. Chem. Education, 7, 782 (1930); Chem. Revs., 8, 191 (1931); J. Chem.

Education, 17, 124 (1940) ; Johnson, J. Chem. Education, 17, 132 (1940).


Page 105: quimica inorganica cuantitativa


In reaction (a) water acts as an acid and OH~ as a base, whereas in (b)

water acts as a base and H30+ as an acid. In reactions (c) and (d) the bi-

carbonate ion acts first as a base and then as an acid.

The classical concept of acids and bases places too much emphasis on

aqueous solutions and obscures the mechanism of many reactions taking

place in nonaqueous solutions. In terms of the Bronsted theory the

mechanism usually appears as both simple and logical. However, the

Bronsted theory does not specifically recognize that ions other than H+are solvated; neither docs it specify the degree of solvation, since this is

unknown. In the calculations of analytical chemistry the use of [H+], the

concentration of hydrogen ion, leads to the same results as would the use

of [H 3O+] or of [(H2O) XII+]; therefore there is little advantage in em-

ploying the concept of the hydronium ion, and its usage is much morecumbersome.


Since neutralization reactions consist essentially of the interaction of

hydrogen and hydroxyl ions, it is important to consider the methods of

expressing the concentration of these ions. It will be recalled from studies

in previous courses in chemistry that the equilibrium constant for anyreversible reaction is derived from the application of the Law of MassAction to the chemical equation for the reaction. It is equal to the productof the concentrations of the substances formed (expressed in moles per

liter) raised to a power represented by the coefficients of the molecules

(or ions), all divided by the product of the concentrations of the reactants

also raised to a power represented by the coefficients. Thus we have for

the reaction

where Ke is known as the equilibrium constant. Since the concentration

of water is constant in any aqueous solution, equation (1) may be written

(2) [H+][OH-] = Ke(R,0] = K.

where Kw is the ionization product for water. Its value changes with a

change in temperature; at 25 it has a value of 10~ 14, and at any constant

temperature its value remains the same whether we are dealing with purewater or with an aqueous solution of an acid, a base or a salt. It follows

then that if the concentration of either ion is known that of the other is

calculable simply by dividing Kw by the known concentration. Obviouslythe acidity or basicity of any solution may be expressed in terms of either

the hydrogen or the hydroxyl ion. For example, a solution of a strong

Page 106: quimica inorganica cuantitativa


monobasic acid which is 0.1 M will yield a concentration of hydrogen ion

of 0.1 M; the concentration of the hydroxyl ion of necessity must be

10"" 14/10~

1 = 10~~ 13. It is logical, even if not customary, to state that such

an acid solution is 10~~ 13 molar with respect to hydroxyl ion.

From equation (2) it also follows that a neutral solution, being one in

which neither the hydrogen nor the hydroxyl ion is in excess, is a solution

in which

(3) [H+] = [OH-] - VH)"'~4 = 10~ 7

and, strictly speaking, neutralization should be defined as a process in

which the resulting concentrations of these two ions have a value of 10~ 7.

It must be remembered, though, that in analyses the object of an acid-

base titration is to bring together chemically equivalent quantities of the

reactants and not, necessarily, to bring about strict neutrality.Instead of expressing the acidity or basicity of a solution in terms of

[H+] or of [OH~] it is customary and more convenient to use a mathe-matical function- of the former known as the pli. The pll is the negative

logarithm of the concentration of the hydrogen ion, or

PH = -log [H+l

Since acidic solutions used in analytical chemistry are seldom greaterthan 1 N the pH of the solutions usually will be represented by small,

positive numbers rather than by the small, fractional numbers whichresult if the hydrogen ion concentration itself is given. This is evident

from Table 7.

Table 7




17VIIC1 . . .

0.1JVHG10.01 JVTTC1. ..

0.00000 1 TV HC1Pure water

0.000001 TVNaOH.01 A^NaOIf.. .


A solution having a pH less than 7.0 is acidic while one having a pHabove 7.0 is basic. The calculation of the pll when the concentration of

*Soreiisn, Biochem. Z., 21, 131 (1909).

Page 107: quimica inorganica cuantitativa


the hydrogen ion is known is quite simple. When the normality of the

solution is an integral power of 10 the pH results at once; it obviously is

the exponent of [H+]with the sign changed as is seen in Table 7. The cal-

culation is never difficult, however, even if sometimes more involved, as

may be seen in the following illustrations involving strong acids and


1. Problem. What is the pH of 0. 15 N solution of HC1 ?

Solution. Since the acid is completely dissociated


]= 0.15 = 1.5 X 10- 1

pll = -log 1.5 X 10- 1

pll = 0.82

2. Problem. What is the pH of a 0.15 N solution of H>S0 4 ?

Solution. [H+] = 0.15 = 1.5 X It)- 1

pll = - log 1.5 X 10- 1

pTI= 0.82

. 3. Problem. What is the pH of a 0.15 M solution of H 2S0 4 P

Solution. Considering the acid as completely ionized and remembering that

one molecule of II 2SOi yields two hydrogen ions

[II*-]= 2 X 0.15 = 3.0 X 10- 1

pll= -log 3.0 X IO" 1

pi I = 0.52

4. Problem. What is the pH of a 0.0050 N solution of NaOH?Solution. Since the base is completely dissociated

[Oil-] = 0.0050 = 5.0 X 10~ 3

By equation (2),

l()-n 10 y TO is

fH+1 = = - = 2.0 X 10-1 J

5.0 X H)~ 3 5.0 X 10~ 3

pll= -

log 2.0 X 10

pll = 11.7


Two factors govern the hydrogen ion concentration established by

solutions of weak acids and bases namely, the concentration and the

degree of ionization. One should recognize at once the difference between

what may be called the analytical concentration of the electrolytic solute,

and the ionic concentration which the solute establishes through partial

dissociation. The analytical concentration represents the total concen-

tration, molecular and ionic, whereas the latter designates only that part

which is in the form of ions.

Consider a solution of a weak monobasic acid, HA, with a concentra-

Page 108: quimica inorganica cuantitativa


tion of C moles per liter. Since it is partly ionized, the molecular HA is in

equilibrium with the ions H+ and A*"

HA i=t H+ + A-

Applying the Law of Mass action we ha\e

[HA]" o

where Ka is the ionization constant for the acid and is a constant for a

given temperature. In a pure solution of the acid evidently there are equalnumbers of H+ ions and A~ ions; therefore

[H+] = [A-]

The concentration of the unionized acid [HA] is equal to the total con-

centration, C, less that which has dissociated, or

[HA] = C -[H+]

Substitution in equation (4) yields




= Ka

Since, however, for weak acids, the value of [H+] is very small com-

pared to C, equation (5) may be written as a close approximation

fH+1 2I-"- J / > if

^ = /v a


(6) [H+] S VC/T

Equation (6) may be employed for weak acids except for very great

dilutions. (At infinite dilution the dissociation is complete even for weak


Problem. What is the pll of a 0.020 N solution of acetic acid? Ka= 1.8 X 10- 5.

Solution. IH+] = V(2-0 X T(F^a.8~X~i6:1


= \/3.6 X 10l7 = V36 X lO- 8


]= 6.0 X 10- 4

pH = -log 6.0 X 10- 4

pH = 3.2

In the case of a weak dibasic acid, H 2A, the ionization takes place in

two steps

H 2A <= H^ + HA-HA 4=i H^ + A-

Page 109: quimica inorganica cuantitativa


but in tlje majority of cases the second ionization step takes place to a

comparatively negligible extent so that the hydrogen ion concentration

may be calculated by equation (6), using K* 19the first ionization constant,

for the dissociation of the dibasic acid. That is,

(6a) [H+]

Now consider a solution of a weak base, BOH, of a concentration of Cmoles per liter. Its ionization is represented by the reversible equation

BOH * B+ + OH-

In a manner exactly analogous to the derivation of the value of [H+]

established by a weak acid, equation (6), it may be shown that

[OH-] ^ VCK,But from equation (2)

[OH-] = /C./IH+]


m -


Problem. What is the pll of a 0.020 /V solution of NH 4OH? Kh - 1.8 X 10" 1.

10-14Solution. [H

f]= -

,_ -----=^- -=V(2.0X10- 2


[11+] = 1.7 X 10-"

pll = - log 1.7 X 10- 11

pll = 10.8

Note that this is above the point of neutrality, 7.0, to the same extent

as the pH for the acetic acid solution was found to be below 7.0. This is

true, of course, because the two factors affecting the pH namely, con-

centration and ionization constants were the same in both cases.


By hydrolysis is meant the reaction of a salt and water to form an acid

and a base. The nature of the solution resulting from hydrolysis will differ

depending upon the kind of salt undergoing the hydrolysis. Therefore, let

us take up the subject from the standpoint of the several types of salts:

salts of strong acids and strong bases, salts of weak acids and strong bases,

and the like. The behavior of salts when they hydrolyze bears importantly

upon acid-base titrations, for when an acid and a base are titrated there

should be present at the end point only salt; the pH at the end point

Page 110: quimica inorganica cuantitativa


therefore is that which is established by the salt formed. An indicator

must be chosen for each titration which will notify the analyst that

chemically equivalent quantities of acid and base have been brought

together. Thus if one knows the pH value which the resulting salt yields,

one may select an indicator which changes color at this same pll value, or

close to it, and therefore may stop the titration at the proper point. Weshall now proceed to see how the pll of aqueous salt solutions is calculated.

Salts of Strong Acids and Strong Bases. A salt of a strong acid and

a strong base, like sodium chloride, does not undergo hydrolysis since

both the acid and the base are completely ionized. Such a solution has a

PH of 7.0.

Salts of Weak Acids and Strong Bases. Consider the sodium salt of

the weak acid, HA. It will react with water,

Na+ + A~ + H 2 ^ Na+ + HA + OH-or essentially

(8) A- + H 2 + HA + 011-

The mass law expression for this reaction will lead to the pH of the salt


[HA][OHJ K[A-][H 20]

But since the value of [H 2O] remains practically constant

where Kh is the hydrolysis constant. Knowing that


and that


we obtain by dividing the former by the latter

[HAJfOH-] = K,*

Evidently then, since Kh of equation (9) and KJKa of equation (10) are

equal to the same thing, it follows that

[HA][OH-] = #j, = K[A"] Ka


Page 111: quimica inorganica cuantitativa


From equation (8) it is seen that for each OH~ ion there is one molecule

of HA. Therefore we may replace [HA] in equation (11) with [OH~] andobtain


_ Kv[A-]" Ka

Assuming complete dissociation for the salt, NaA, so that [A~] =

[OH~], where C is the concentration of the salt in moles per liter, and

remembering that [OH~] is much smaller than C, we have

[o?2 _ JK.

C =Ka


Since from equation (2)

[OH-] =then

. .

- V K.from which


Problem. What is the pH of a 0.025 /V solution of


X 10~ 18

[H+]= 2.7 X 10~ 9

pH = -log 2.7 X 10-'

pH = 8.6

Salts of Weak Bases and Strong Acids. The corresponding case of

the pH of a solution of a salt of a weak base and a strong acid is even

simpler. Consider the chloride salt of the weak base, BOH. It will react

with water

B+ + Cl- + H 2 <= BOH + H+ + Cl-

or essentially

B+ + H2 < BOH + H+

From the Law of Mass Action and assuming the concentration of water

to be constant, we obtain

[BOHHH+1 _ * *ru-4-i ^^ TS

Page 112: quimica inorganica cuantitativa


Kb being the ionization constant of the weak base. Now [BOH] = [H+]

and [B+

] S C, where C is the concentration of the salt, and substituting

these values in equation (13)


from which


Problem. What is the pH of a 0.025 N solution of NH 4C1?

Solution. [H+]=

(10~14)(2.5 X 10-2


1.8 X 10-*

= V1.4 X 10- 11


]= 3.7 X 10-*

pH = -log 3.7 X 10-'

PH = 5.4

Salts of Weak Acids and Weak Bases. In the two former cases the

pH established by the salt was found to be governed by the concentration

C of the salt and by the ionization constant Ka or Kb . In the present case

we shall see that the ionization constants of both the weak acid and weak

base (from which the salt is regarded as being formed) influence the pHof the salt solution. The salt of the weak acid, HA, and the weak base,

BOH, may be used to illustrate this type of reaction. The ions from this

salt, BA, will react with water as follows

B+ + H2O ^ BOH + H+and A- + H2 * HA + OH~,

from which it follows that

[BOHHH+] K.(15)


- A. -Kt


/,[HAKOH-] _ K =



Of course, simultaneously, the following also must be satisfied:

Dividing the product of the two former by the last yields

KktK* Kw


Page 113: quimica inorganica cuantitativa


Assuming complete ionization of the salt and that the weak acid and the

weak base are ionized to about the same degree, it follows that

[HA] s [BOH]and [B+] S [A~] S C

Substituting these values in equation (17)

C 2=


Substitution of this value of [HA] (and also C for [A"]) in the mass law

expression for the ionization of the weak acid, HA, namely,

[H+KA-] _ K -



will yield the value of [H+], thus:



or, solving for [H+],


Note that the following conclusions may be drawn regarding the

solution of a salt of a weak acid and a weak base. First, the pH is inde-

pendent of the concentration of the solution. Second, from equation (17)

it is obvious that the smaller the value of the product KaKb the larger

will be the value of Kw/KaKb and thus the greater the extent of the

hydrolysis. Third, from equation (18) it is evident that when the ratio of

Ka/Kb exceeds unity (that is, when the ionization constant of the weakacid is greater than that of the weak base), the solution of the salt will

give an acidic reaction; the pH will be less than 7.0. Conversely, if the

ratio Ka/Kb is less than unity the solution of the salt will give a basic

reaction; the pH will be greater than 7.0. A salt like ammonium acetate

will give a neutral reaction since the values of Ka and Kb for acetic acid

and ammonium hydroxide are almost identical, 1.8 X 10~6.

Problem. What is the pH of a solution


1.8 X 10- B

H+]= 10-7

pH = -log 10- 7

pH = 7.0

Page 114: quimica inorganica cuantitativa


Problem. What is the pH of a solution of HCOONH4 ? Ka = 1.8 X 10~ 4;

Kb= 1.8 X 10-*.

Solution. [H+]


1.8 X 10-*

[H+] = 3.2 X 10~ 7

pH = -log 3.2 X 10- 7

pH = 6.5

Acid-Salts. Consider the primary sodium salt of the dibasic acid H2A.

In solution it ionizes according to the equation

NallA - Na+ + HA~

But the anion HA~ itself will ionize to some extent:

(19) HA- < H+ + A-and also will hydrolyze:

(20) HA- + H 2 <= H 2A + OH-

The reaction of equation (19) yields H+ ions while that of (20) yields OH~ions. If we assume, as is true in most cases, that these two reactions go

forward to about the same extent, then

(21) [A=] s* [H 2A]

The mass law expressions for equations (19) and (20) are




and dividing the former by the latter

[H2Af[OH-] Kw

From equation (21) we may substitute [H 2A] for [A=

] and from equation

(2) we may replace [OH~] by KW/[H+]. Making these substitutions in

equation (22) and solving for [H+] gives

(23) [i-i+] ^ VK^K!,

This approximation may be employed only if Ka,and Kat are small com-

pared with C, the concentration of the salt; otherwise equation (21) is

invalid and thus also equation (23).

Problem. What is the pH of a solution of NaHCO 3 of moderate concentration?

Ka,= 3.0 X 10-7

; Ka,= 6.0 X 10-".

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Solution. [H+] = \/(3.0 X 1Q- 7)(6.0 X 10~n)

[H+] = 4.2 X 10- 9

pH - -log 4.2 X 10-*

pH = 8.4

Normal Salts of Dibasic Acids. The equation for the hydrogen ion

concentration of a solution of the normal, or secondary, salt of a dibasic

acid is derived in a manner analogous to that of equation (12) except that

Ka is replaced by Kaz , the secondary ionization constant. That is,


Problem. What is the pH of a 0.0100 M solution of Na 2CO 3 ? K^ = 6.0 X

X 10-J

[H+] = 7.7 X 10- 12

pH -- log 7.7 X 10- 12

pH = 11.1

Buffer Solutions. A solution containing both a weak acid and a salt

of that acid, or one containing both a weak base and a salt of that base,

will establish a pH which is not greatly affected by the addition of

moderate amounts of even strong acids or bases. Such solutions arc called

buffer solutions. Often in quantitative analyses provision must be made to

keep the pH of a solution fairly constant. The use of a buffer is generally

employed for this purpose. The particular pH which a certain buffer will

give depends upon two factors: the concentrations of the electrolytes

employed and the ionization constant of the weak acid (or weak base).

Therefore, within reasonable limits, a buffer of any desired pH may be

prepared.The hydrogen ion concentration of a buffer made with a weak acid

and one of its salts may be calculated from an equation developed as

follows. The mass law expression for a weak acid, HA, and its ions is


[H+] =^ Ka

In the presence of one of its salts the value of [A~] will be approximatelythat furnished by the salt alone, or C, since the acid is weak and the salt

is assumed completely ionized. Thus the above equation becomes

(25) |HA J = p **-a

A buffer made up of a weak base and one of its salts will set up a

hydroxyl ion concentration found, by a derivation similar to that of

Page 116: quimica inorganica cuantitativa


equation (25), to be

or, substituting KW/[H+] for [OH-] and solving for [H+],

(26) 1H+] =

Equations (25) and (26) therefore may be used to compute the value

of [H+], and thus the pH, of all such buffer solutions. Note that in both

cases the value of [H+] is equal to a constant (Ka for acid buffers and

Kw/Kb for basic buffers) multiplied by the ratio of the concentrations of

the acid and the salt (or of the salt and the base). Since the acid (or the

base) and its salt are in the same solution, the volume of the solution has

no effect upon the hydrogen ion concentration, unless, indeed, the dilu-

tion be extremely great.

The reason that a buffer solution resists changes in pH lies in the fact

that no matter whether an acid or a base is added to the buffer a reaction

takes place to prevent most of the newly added 11+ ion or OH~ ion from

remaining as such. An acid buffer may be used to illustrate the point. If

acetic acid and sodium acetate make up the buffer, the solution contains

acetic acid molecules and hydrogen, sodium, hydroxyl and acetate ions.

If hydrochloric acid is now added the hydrogen ions from it largely com-

bine with the acetate ions to form more acetic acid molecules, and the

value of [H+] varies only slightly. On the other hand, if sodium hydroxide

is added to the buffer the hydroxyl ions from it largely combine with

hydrogen ions to form water, and again the value of [H+ ]is changed very

little. A buffer, to be most effective that is, to resist changes in hydrogen

ion concentration when either acid or base is added should be made up

in an equimolar ratio of weak acid (or weak base) and salt, for then

potentially there will be present sufficient concentrations of ions to take

care of either hydrogen ion or hydroxyl ion from the newly added acid or

base. This refers, of course, to the capacity of the buffer to offset changes

in pH. The particular pH which a given buffer sets up and approximately

maintains is governed, let it be repeated, by the magnitude of the ioniza-

tion constant involved and by the concentrations of the acid (or base)

and its salt.

Suppose that a buffer having a pH of about 5 is wanted. A weak acid

having an ionization constant of around 10~ 5 may be employed. From

equation (25) it follows that equimolar quantities of acetic acid and

sodium acetate in solution will give a pH of 4.74 for

[H+] = K(1.8 X 10- fi


pH = -log 1.8 X 10-5

pH - 4.74

Page 117: quimica inorganica cuantitativa



If a buffer of exactly pH = 5.0 is desired, acetic acid and sodiumacetate may still be used but not in equirnolar ratio. The proper ratio in

this case is calculated from equation (25). For a pH of 5.0 the value of

[H+] is 10- 5. Therefore

10-* =t-p^ (1.8 X 10-*)

[HAc] = J_

That is, if a solution is made up containing 1 mole of acetic acid to 1.8

moles of sodium acetate it will give a pH of 5.0.

To illustrate the power of buffers to resist variation in pll we maycalculate the change brought about by adding either a strong acid or a

strong base to an unbuffered solution on the one hand, and to a buffered

solution on the other.

Suppose we have 100 ml. of pure water (or of any neutral unbuffered

solution). The pH will be 7.0. If the water contained 5 ml. of exactly 0,1

M HC1 the resulting solution obviously would be 0.0050 M with respectto HC1 and

[H+] = 5.0 X 10-3

pH = 2.3 and ApH = 4.7

If, instead of HC1, the water contained 5 ml. of exactly 0.1 M NaOH,then

FH+1 = Kw = 10~ 14

1 J

[OH-] 5.0 X 10~3

[H+-] = 2.0 X 10-"

pH = 11.7 and again ApH = 4.7

Suppose, on the other hand, we make up a buffer by adding together

exactly 50 ml. of 0.1 M HC 2H 3O 2 and 50 ml. of 0.1 M NaC 2H 3O 2 . As has

been pointed out already, this buffer will have a pH of 4.74, since from

equation (25)



pH = 4.74

If the 100 ml. of buffer also contained 5 ml. of 0.1 M HC1 the hydrogenions of the hydrochloric acid would combine almost completely with

acetate ions of the buffer, thus bringing the value of [HAc] to 0.0550 and

simultaneously decreasing the value of C to 0.0450. Therefore

[R+] =0.0550

in J0.0450 u ' X 1U ;

[H+] = 2.20 X 10-

pH = 4.66 and ApH = O.OR

_ .


i J""

(1 'b X 10 >

Page 118: quimica inorganica cuantitativa


If, instead of hydrochloric acid, the buffer contained 5 ml. of 0.1 MNaOH, the hydroxyl ions would combine almost completely with hy-

drogen ions of the buffer, thus decreasing the value of [HAc] to 0.0450 and

simultaneously increasing the value of C to 0.0550. Therefore

-8 >< w-'

[H+] = 1.47 X 10-5

pH = 4.83 and ApH = 0.09

Thus, whereas the addition of 5 ml. of 0.1 M solution of strong acid or

base to 100 ml. of unbuffered solution changed the pll by 4.7 pH units,

the same treatment of the acetic acid-acetate buffer changed the pH byless than 0.1 pH unit. That is to say that the effect of the acid or base

proves to be, in the case at hand, about 50 times as great upon the un-

buffered solution as upon the buffered solution (4.7/ < 0.1 = 50).


Acid-base indicators are organic dyes which exhibit the characteristic

of changing from one color to another, or becoming colorless, when the pHof the solution in which they are dissolved is varied slightly. The particu-

lar pi I value at which the change in color occurs is different for different

indicators. Therefore an indicator, if properly chosen, may be employedto mark the end point for an acid-base titration, provided the pH estab-

lished when chemically equivalent quantities of acid and base have been

brought together is known. Indicators are usually weak organic acids or

bases and their tinctorial power is so high that they may be used in very

low concentration. The color which a given indicator assumes at any pHvalue may be ascertained by placing a few drops in each of a series of

buffer solutions of definite and ascending pH values. Methyl orange indi-

cator, for example, in an acidic solution with a pH of 1 shows a pink color.

In a solution of a somewhat higher pH the methyl orange begins to show

an orange color. The orange tint first becomes apparent at a pH of about

3.1. After the pink color has completely disappeared and is replaced by

pure orange the pH will be found to have a value of around 4.4. The pHinterval, 3.1 to 4.4, during which the color was changing from pink to

orange, is called the range of the indicator. Within the range of the indi-

cator there is a pH value at which a distinct and fairly abrupt change of

hue takes place; this happens ordinarily when the tint is about midwaybetween the two extremes of color, and in the case of methyl orange at a

pH of about 4.0. This is called the transition pH of the indicator.

Table 8 gives a few of the indicators commonly used in volumetric

analysis and the pH at which the color change occurs.

Page 119: quimica inorganica cuantitativa


Table 8


* Data from W. M. Clark, The Determination of Hydrogen Ions, Williams & Wilkins Co., Balti-

more, 1928.

The transition color of two-color indicators is somewhat a matter of

individual opinion. One analyst may consistently carry his titrations to a

slightly higher pH than another when using methyl orange, for example;but so long as he is consistent in obtaining the same tint of orange color

every time, it is not important that the exact hue to which he titrates is

slightly different from that obtained by another worker.


Since the organic dyes used as indicators for acid-base titrations are

either weak acids or weak bases the reaction for their ionization may be


HIn <=* H+ + In-

InOII * In+ + OH-(27)


Consider the first, a weak acid indicator, as an example. Ostwald 3

suggested that such an indicator owes its color change to the fact that the

molecule, HIn, is of one color (or colorless, as in the case of phenol-

phthalein) and the anion, In~% is of another color. Therefore an acid indi-

cator in an acidic environment should be, for all practical purposes, un-

dissociated; that is, the ionic equilibrium should be shifted so far to the

left (equation (27)) that practically none of the colored In~ ion remains

in the solution. In a basic environment the indicator reacts with the base

of the titration (after the acid of the titration has been satisfied) formingthe salt e.g., Naln which salt is highly ionized and therefore impartsthe characteristic In~~ color to the solution.

There is much evidence that the mechanism involved when indicators undergoa color change is not so simple as depicted above. The organic chemist long since

has established through his elegant syntheses that color is associated with definite

3 Ostwald, Die wissenschaftlichen Grundlagen der analyiiche Chemie, W. Engelmann,Leipzig, 1894.

Page 120: quimica inorganica cuantitativa


molecular structure. At least one of these color groups within the organic molecule

is always present when color manifests itself; such groups are called chromophores.Reduced to its simplest argument it is thought that indicators change color, not

because of ionization, per se, but because the ionization is accompanied by intra-

molecular rearrangement which either produces or eliminates a chromophoric

group. This point of view may be illustrated with what is thought to take place

in the case of phenolphthalein.

The quirioid chromophore, =\ />= , is not present in the structure of>- ' * /

the phenolphthalein molecule, and the molecule is colorless. In the presence of a

small or a moderate amount of excess base, phenolphthalein is converted into an

unstable intermediate which loses water to form the disodium salt. The anion of

this salt does contain the chromophore and the red color appears. The fact that a

large excess of base causes phenolphthalein to lose its red color is accounted for bythe formation of the trisodium salt in which the quinoid structure has been lost.

These relationships are shown in the following structural formulas.


C O Na




VAnion of

disodium salt (red)





Anion of

trisodium salt (colorless)

Only the anion of the disodium salt of phenolphthalein contains the chromophoric

function, =z\ /^ therefore it alone gives the red color.

Page 121: quimica inorganica cuantitativa


The interpretation given by Ostwald to account for color changes in

indicators is obviously simpler than the chromoplioric theory and, while

the latter is doubtless nearer the truth, we may employ the more ele-

mentary approach in studying the ionic equilibrium involved. Applyingthe mass law to equation (27) we obtain

ran [H+][In-] _(28)



(29) [H+] = [ 5l K*

where K Jn is the indicator constant. The intensity of color is proportionalto the concentration of the colored unit, so that for a two-color indicator

equation (29) may be written

Intensity of Color 1,on x ru-H(60) 1*1 J T

- ----rz--

f r* i TTIntensity of Color II

from which it follows that the color actually seen depends only upon the

hydrogen ion concentration. Now the eye has its limits in recognizing a

given color when two colors are present. If, as in the case of methyl orange,

the colors are red and yellow, the eye is unable to distinguish between purered color and that resulting from a mixture of 95 per cent red and 5 percent yellow. As a matter of fact the uncertainty generally exists until the

ratio of [HIn] to [Iri~] is about 10. In a solution in which the ratio is greater

than 10 the eye gets the impression that the color is that of the pure"acid" hue; likewise, if the ratio is less than 0.1 the color appears to be

that of the pure "basic" tint. The range of any indicator is that pH spanfor which the above-mentioned ratio is between 10:1 and 1:10. Between

these ratios the eye registers an intermediate color. It follows that an

indicator changes its color abruptly only if the pH suddenly varies over

the range of the indicator. If we substitute the values 10 and 0.1 for

[HIn]/[In] in equation (29) we obtain [H+] = 10#In and [H+] = O.ltfln ,

or in terms of pHpH = pKln 1

where pKin= log Kin . Thus the pH range necessary for a change from

one color to the other for a given indicator will be

ApH =(ptfln + 1)


-1)= 2

The above calculation of ApH is based upon the assumption that the one

colored form must be of a concentration 10 times the other before the eye

registers the effect of a pure color. The confirmation of the validity of this

assumption may be seen in Table 8 where the range for typical indicators

is found to be about two pH units. Finally, it should be noted that if,

during a Nitration, the concentration of the reagent added from the buret

Page 122: quimica inorganica cuantitativa


is such that one <lyop will cause a change of two pH units in the solution

being titrated, a sharp end point will result. If, however, two or three

drops must be added to produce a change in pll of two units, the end

point will be gradual and the titration less satisfactory.

Mixed Indicators. It is sometimes possible to secure a sharper and a

more pronounced color change by using together two indicators or oneindicator and a dye. For example,

4 instead of methyl orange, one may use

a solution of 1 g. of methyl orange and 2.5 g. of indigo carmine per liter

of water. This gives a gray color at a pH of 4.0, but is violet below 4.0

and green above 4.0. Again 0.7 g. of methylene green and 0.3 g. of phenol-

phthalein in a liter of solution gives a mixed indicator which is light blue

at a pH of 8.8 and is respectively green and violet immediately below andabove 8.8.


It has already born stated that the object of any titration is to bring

together chemically equivalent quantities of mutants. Furthermore, wehave seen that when this is done in the case of an acid and a base the

resulting salt solution may have a pH above, below or equal to 7.0 de-

pending on the particular titration. We know also that different indicators

have different pH ranges. From these facts it should be clear that the

proper indicator for a given titration is that indicator having a pH rangewhich includes the pll established by the salt formed as a result of the acid-

base lilralion. Any indicator meeting this requirement will be suitable pro-vided that its pH range is spanned upon the addition of one or two dropsof reagent from the buret. The pH which is established by the salt result-

ing from the titration is calculated, as already set forth, from equation

(12), (14), (18), (23) or (24). This done, the proper indicator may be

chosen from Table 8 or from more extensive tables found elsewhere.


The manner in which the pll varies during the course of a titration is

best displayed by plotting a graph with the pll values as the ordinate andthe volume of reagent added from the buret as the abscissa. Such curves

may be constructed from pH measurements obtained experimentally, or

the data for the graphs may be calculated. The calculated curves are in

general a close approximation of those plotted from experimental data;the fact that they are approximations is understood when it is remem-bered that most of the equations already developed for computing pHvalues involve simplifying assumptions. In deriving these equations the

point at which the approxinfation symbol, =^, first appears denotes the

introduction of the slight departure from strict equality.

4 See also Kolthoff and Stenger, Volumetric Analysis, Vol. II, Interscience Pub-Ushers, New York71$4t.

Page 123: quimica inorganica cuantitativa


It must be pointed out, furthermore, that we have defined pH as equal to

log [H+] whereas as a matter of fact pH = log (11+) where (H+) is the

activity of the hydrogen ion. In general, calculations which are involved in ionic

equilibria are not quite in accord with experimental observations when the con-

centrations of ions are employed. Deviations become more pronounced as the

solutions become more concentrated, and especially in solutions containing

foreign ions. Debye and Hiickel 6 have shown the relationship between concentra-

tion and activity based upon the assumption that solutions of strong electrolytesare totally ionized and that there exists an attraction between ions due to the

electrical charges which they bear. The factor by which the concentration mustbe multiplied in order that such calculations as are involved in solubility products,

pH, etc., may tally precisely with experimental results, is called the activity

coefficient. The coefficient thus is a factor which gives the fraction of the concen-

tration of an ion which is effective in the equilibrium under consideration; that is,

it measures the effective ioiiization. At infinite dilution the activity coefficient

becomes unity but at moderate concentrations the coefficient for a given ion in a

given solution will be significantly less than unity. In regard to pH with which weare at present concerned the error caused by using [I I'


] instead of (II +) usuallyamounts to less than 0.2 t>H unit.

The calculation of titration curves becomes a simple matter when it is

borne in mind that there afe just four types of points to be computed.

They are: (1) the pH at the beginning, before any reagent has been added

from the buret; (2) the pH for points after additions from the buret but

before the stoichiometric point; (3) the pH at the stoichiometric point;

and (4) the pH for points past the stoichiometric point.

The methods of calculating these four types of points are somewhatdifferent for the titration of a strong acid and strong base from the

methods used when one of the reactants is weak. We now may proceed to

illustrate the calculations for a few typical titrations.

Titration of a Strong Acid by a Strong Base. Consider the titra-

tion of 30 ml. of 0.1 N HC1 by 0.1 N NaOH. Suppose that first the acid is

diluted to a total volume of 100 ml. As the titration proceeds the total

volume, of course, increases. The calculated values for the pH at various

stages of the titration are given in Table 9. These should be checked bythe student. The calculations of four typical points are given below. 6

1. Before the addition of any base the solution consists only of hydro-chloric acid, completely dissociated. The 0.1 N HC1 (= 0.1 M) wasdiluted to 100 ml., therefore

[H+] = = 0.0300 = 3.00 X

pH = -log 3.00 X 10- 2 = 1.52

5 Dcbye and Hiickel, Physik. ., 24, 185 (1923). See also Clark, The Determination

of Hydrogen Ions, Williams and Wilkins Co., Baltimore, 1928, Chap. XXV; Briscoe,General Chemistry for Colleges, Houghton Mifllin Co., New York, 1943.

6 In the following calculations it will be assumed that volumes are accurate to 0.01

ml. and concentrations to 0.0001 M so that the pH may be calculated with a precisionof 1 part per 1000 if desired.

Page 124: quimica inorganica cuantitativa


Table 9



2. After the addition of 20 ml. of base the solution may be thought of

as containing 10 ml. of 0.1 TV free acid plus 20 ml. of 0.1 TV salt, diluted to a

total volume of 120 ml. The salt does not appreciably affect the ionization

of the HC1, therefore

[H - 8.3.3 X 10-

pH = -log 8.33 X 10- 3 = 2.08

3. At the stoichiometric point the solution contains only NaCl. This

is u salt of a strong acid and a strong base and therefore does not hydro-

lyze; thus the pH is 7.00.

4. When excess base has been added the solution contains salt andexcess hydroxyl ion. The salt has no great effect upon the concentration

of hydroxyl ion which therefore depends upon the quantity of base added.

For example, when 40 ml. of NaOH has been added we may regard the

solution as containing salt and 10 ml. of 0.1 TV free base diluted to a total

gf 140 ml. Therefore

Page 125: quimica inorganica cuantitativa


[OH-] =

pH = 11.85

The data of Table 9 are plotted as a solid curve in Fig. 20. The shaded

areas represent the range of the two indicators, phenolphthalein and

methyl orange, as taken from Table 8. It is clear that because the steep

portion of the curve extends over a long range in pH either phenol-

phthalein or methyl orange may be used as the indicator. Theoretically if

phenolphthalein were used the end point would come when about 30.03

ml. of base have been added, for the pll would then be high enough to

come within the color range of the indicator. On the other hand, if methyl

orange were used the end point theoretically would come at about 29.95

ml. of base (if the titration be carried to pure yellow), for then the pH is

4.41 which is just to the higher side of the range of this indicator. The

difference in using the two indicators should be about 0.1 ml. of base;

in actual practice if 0.1 N solutions are used the difference found by most

analysts amounts to about 0.15 ml. of base.

Obviously in secondary standardizations of strong acids against strong bases

two values for the normality of a solution may he obtained, one from data secured

by titrating when phenolphthalein is used and another when methyl orange is

used. Suppose the values obtained are, respectively, 0.1035 Wand 0.1029 N. The

question sometimes is asked, which is the correct normality? The answer is, of

course, that each is correct. \Vhcn this solution is used in future analyses the

value 0.1035 should be employed if phenolphthalein is the indicator and 0.1029

should be employed if methyl orange is the indicator. In other words, we only

need to be consistent.

Fig. 20 also gives a second curve, for the titration of 0.001 N HC1 by0.001 N NaOH. By calculations exactly analogous to those from which the

solid curve for 0.1 N solutions was plotted it may be seen that the pH after

the addition of a given quantity of base is exactly 2.00 pH units higher for

the dilute solutions until the stoichiometric point is reached; there the

curves coincide. For volumes of base past the stoichiometric point the pHfor the dilute solutions are 2.00 pH units lower than that for the 0.1 Nsolutions. This difference of 2.00 pH units must occur since the concen-

trations differ in the two cases by 10 2, the logarithm of which is 2.00.

The more dilute the solutions the shorter is the steep portion of the titra-

tion curve. For 0.001 N solutions it is evident that neither phenol-

phthalein nor methyl orange could function as the indicator; instead an

indicator with a color change taking place at a pH of between 6 and 8 is

necessary. The kinds of difficulties encountered if one were to use either

phenolphthalein or methyl orange for such a titration, if not already

foreseen, will be clarified in the following section.

Page 126: quimica inorganica cuantitativa


The flat portion of the curve in Fig. 20 from to 25 ml. is not usually

said to be due to buffer effect. It may be so called if a buffer solution were

defined as any solution which resists change in pH. If, however, a buffer

is denned as a solution of a weak acid (or base) and a salt of that acid (or

base), then the titration of HC1 by NaOH involves no buffer action.

Rather, the reason for the flat portion of the curve is that here the solu-

tion is so strongly acidic that the small increments of base do not greatly

vary the pH.






' I

10 15 20 25 30

Ml. NaOH

35 40 45

FIG. 20. Titration curves for HC1. Curve 1, 0.1 N I1C1 with 0.1 N NaOH;Curve 2, 0.001 N HC1 with 0.001 N NaOH. Diluted from 30 ml. to 100 ml. before


Page 127: quimica inorganica cuantitativa


Titration of a Weak Acid by a Strong Base. The construction of

the titration curve for a weak acid and a strong base is somewhat more

involved than that of the case first considered for three reasons: the acid,

being weak, is incompletely dissociated; as the titration proceeds a salt

is formed which represses the ionization of the acid; and the salt hydro-

lyzes. The first of these factors, incomplete dissociation of the acid, in-

fluences the calculation of the pH before any base has been added. Thesecond affects calculations for points after addition of base but before the

stoichiometric point is reached. The third influences the pH at, and im-

mediately before and immediately after, the stoichiometric point. Thetitration of acetic acid by sodium hydroxide will illustrate this type of

reaction. Table 10 lists pH values calculated for the titration of 30

ml. of O.I TV HAc first diluted to 100 ml. and titrated with 0.1 N NaOH.

Table 10


Typical points are calculated as follows:

1. Before the addition of any base the solution contains only HAc.

The concentration of the acid is

Page 128: quimica inorganica cuantitativa


= 3.00 X 10-' moles per liter

If we let X = [H+] then X = [Ac-] and

__ __lz. _ i v in *

(3.00~X 10" 2)- X

" 1-* X W

Throwing out 7 X from the denominator as insignificant

X 2 = 54 X 10- 8

X - 7.35 X 10~ 4

pH - -log 7.35 X 10-4

pH = 3.13

2. As an example of the calculation of a typical point after the titra-

tion has begun but before the stoichiometric point has been reached, sup-

pose that 20 ml. of base has been added. Here we have free weak acid

present plus salt of that acid; thus we have a buffer solution. Accordingly

equation (25) applies for the calculation. We may think of the solution

as consisting of 10 ml. of 0.1 N (= 0.1 M) free acid and 20 ml. of 0.1 Nsalt, diluted to a volume of 120 ml. Therefore,




t1 -8 X

120= l Ho(l.* X 10- 5


[H+] = 9.00 X 10- 6

pH = -log 9.00 X 10~ 6 = 5,05

7 If X is not discarded we have a quadratic equation:

X 2 + 1.8 X 10" 6X -(5.1 X 10~7


solved by applying the formula

\ -b Vb 2~lac2a

Solving for X by the quadratic formula we find the value to be identical to two signifi-cant figures with the value obtained by the approximation. As a rule in such calcula-tions the approximation is not valid if solving after discarding X to the first poweryields a value for X which is appreciable in comparison with the term to or from which,in the original equation, it is added or subtracted. To illustrate, in


0. 2 -X=-2

if the X is thrown out we have as an "approximation"X2 - (0.2) (0.2)

X = 0.2

whereas actually X - 0.124 (by quadratic solution). Therefore in this case, since theterm from which X was subtracted in the original equation, namely 0.2, is appreciablein comparison with the value obtained by the approximation, also 0.2, we must solve

by the exact method.

Page 129: quimica inorganica cuantitativa


and similarly, after the addition of 25 ml. of base,

[H+] = ^5(1.8 X 10-')

= 3.60 X 10- 6

pH = -log 3.60 X 10-6 = 5.44

Equation (25), namely, [11+] ^ p Ka , it will be remembered, is only an

approximation; it ignores the fact that [HA] is somewhat decreased and C is some-

what increased due to the slight ionization of the weak acid. Thus the true value of

the numerator in equation (25) should be decreased to an extent governed by the

degree of dissociation, and the denominator should be increased to an identical

extent. The exact expression thus becomes, if we let X equal the dissociation of the

acid in moles per liter, and thus also equal [H+],

v [HA] - X v


It is instructive to see just when equation (25) suffices and when equation (25a)

is necessary. We may do this by calculating several points for Fig. 21 by both

equations. If we do so the following values are obtained.

Table 11


To illustrate the calculations by which the figures of the third column of Table

11 are obtained one such calculation is given. After 1.000 ml. of base has been

added we have 29.00 ml. of acid remaining and also have 1.000 ml. of salt. Thetotal volume of the solution is 101.0 ml. Therefore

[HA] =


C =




= 2.87 X 10-2

= 9.90 X 10- 4

Substituting in equation (25a)

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_ [(9.90X 10-4

) + (1.8 X 10-') I

X 10-.) + (1.8 X 10-,|- x , x ,.

4= -5.04 X 10- 4 + 8.78 X 1Q- 4

X = [H+] = 3.74 X 10- 4

pH = 3.43

It is seen from Table 11 that after 5 ml. of base has been added the exact equa-tion (25a) is no longer needed. This, of course, is due to the fact that we are deal-

ing with buffer solutions with a common ion effect and in the early stages of the

titration the solution contains very little salt; therefore here the value of C in

equation (25) is increased to a relatively great degree by the anion coming fromthe weakly dissociated acid. Likewise, when there is still little salt formed, the

repression of the ionization of the acid is negligible, so that [HA] in equation (25)

here must be significantly decreased, by the extent of its dissociation. Both of

these corrections are made in equation (25a). But as larger quantities of base

are added the amount of salt present increases, with the result that C more nearly

approaches the value of the anion supplied by the salt alone, and the value of

[HA] becomes (due to the common ion effect) more nearly that of the free acid

alone. The net result of the exact calculation for the first few points in Fig. 21 is a

steeper initial rise of the curve which, around 5 ml. of base in the example cited,

gives way to a smaller slope (flatter curve) as the effectiveness of the buffer action

becomes more pronounced.It should be pointed out here that this same situation prevails for the initial

stages of the titration when the weak acid is titrated with a weak base instead of

with a strong base. If acetic acid is titrated with ammonium hydroxide we have

present as the titration proceeds free acid arid the salt, ammonium acetate,

whereas with sodium hydroxide as the base we have present free acid and the

salt, sodium acetate. In both instances then the components present in this regionof the curves are a weak acid and a salt of the acid.

3. At the stoichiometric point equation (12) is used. The value of C is

evidently3% 3 o(0.l) = 0.023, so that


2.3 X lO- 2

[H+] = 2.80 X 10- 9

pH = -log 2.80 X 10~ 9 = 8.55

4. When excess base has been added the solution contains NaOH andNaAc. The hydrolysis of the salt thus is repressed and the pH is calculated

exactly as for the corresponding point for the titration of a strong acid

by a strong base. That is to say, the titration curve for a weak acid-

strong base titration coincides with the curve for a strong acid-strongbase titration after the stoichiometric point has been passed (base in the

buret). This obviously must be true since in both cases the solution con-

tains free base plus a salt: NaOH and NaAc in the one case, and NaOHand NaCl in the other.

Page 131: quimica inorganica cuantitativa


20 25 30

Ml. NaOH

FIG. 21. Titration curves for HC 2H 3 2 and H 3B0 3 with O.I W NaOH. Curve 1 ,

0.1 N HCsH 3 2 diluted from 30 ml. to 100 ml. before titrating; Curve 2, H 3B0 3 .

Note the rather pronounced change in slope of Curve 1 at about 5 ml. of

NaOH (contrast this with the HCl-NaOH curve of Fig. 20 at 5 ml. of NaOH)due to the repression of the ionization of the weak acid by the salt, NaC 2H 3 2 ,

which by now has been formed in considerable quantity. From this point on to

about 25 ml. of NaOH, both weak acid and salt of that acid are present in fair

amounts; as a result the solution is buffered in this region. From 25 ml. to 30 ml.

of base the buffer action again is very slight since only small amounts of acid

remain, and the pH resumes its rapid rise with small increments of base.'

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The titration curve for acetic acid, Fig. 21, shows that, of the three

indicators, phenolphthalein, methyl red and methyl orange, only the

first may be used to detect the end point for the titration. Consider what

would happen if one attempted the use of methyl orange. Since before any

base has been added the solution of acetic acid already has a pH of 3.13,

the color of the indicator, while predominantly pink, will be tinged with

yellow. As base is added from the buret the color will gradually become

more nearly pure yellow but the slow change in color will be extended over

the addition of sodium hydroxide to about 10 ml. Thus there is no abrupt

color change and even when the pure yellow does appear at a pH of about

4.4 only 30 per cent of the base necessary for chemical equivalence has been

added. The titration error would amount to 70 per cent or 700 parts per

1000. Methyl red will not do either for it begins to change color at a pHof about 4.2 and becomes pure yellow when only 95 per cent of the theo-

retically equivalent quantity of base has been added. Thus a titration

error of about 50 parts per 1000 results when it is used. Therefore, for

such a weak acid as acetic acid an indicator like phenolphthalein with a

narrow pH range occurring between 7 and 10.5 (the steep portion of the

curve) must be employed. To summarize, methyl orange and methyl red

fail as indicators for a weak acid-strong base titration because (a) the

color change is not abrupt, and (b) even when the pure color is obtained

only a fraction of the quantity of base necessary for chemical equivalence

has been added. With phenolphthalein, however, the end point is sharp

and is practically coincident with the stoichiometric point. Thymol-

phthalein also is satisfactory.

Titration of a Weak Base by a Strong Acid. The calculation of the

points for the titration curve of a weak base-strong acid titration again

involves the four types of points. Suppose 30 ml. of 0.1 /V NH 4OH diluted

to 100 ml. is titrated with 0.1 N HC1.

I. Before the addition of any acid from the buret the solution con-

tains excess hydroxyl ion. The concentration of the base is

- 3.00 X 10--

Tf X = [OH-] then X = [NH 4+] and


X = 7.35 X 10-4

since [H+] = A./[OH-]10~ 14^ -


- L36 X

pH = -log 1.36 X 10-"

pH - 10.87

Page 133: quimica inorganica cuantitativa


2. After the addition of HC1 but before the stoichiometric point hasbeen reached, for example, when 15 ml. of HC1 has been added, fromequation (26) we have

rw+l = (15) (10-")1 j

(15) (1.8 X 10- 6)

= 5.56 X 10- 10

pH = -log 5.56 X 10~ l

pH = 9.26

All that was said in regard to replacing buffer equation (25) with equation(25a) in calculating the exact pH for points up to 5 ml. of base in the titration ofacetic acid with a strong base, must be borne in mind in constructing the presenttitration curve. Thus for the first several points for the ammonium hydroxide

curve (up to 5 ml. of HC1) instead of equation (26), namely, [Hf

] & --,


one should use the more exact equation

(26a) X = - (^^ J + J^ V" +

and [H+] = ^where X represents [OH~]. We arrive at equation (26a) by reasoning exactlyanalogous to that which leads to (25a). For comparison, after the addition of 1 ml.of HC1 in our titration of ammonium hydroxide, equation (26) yields an approxi-mate pH value of 10.72, whereas by the exact calculation the pH equals 10.57, adifference of 0.15 pH unit.

3. At the stoichiometric point equation (14) is used. The value of C is

obviously39iso(0.1) or 2.30 X 10~ 2 so that


]=V ir^rio^

= 3.58 X 10- 6

pH = -log 3.58 X 10- 6

pH = 5.45

4. When excess acid has been added, say 31 ml., the calculation of the

pH is analogous to corresponding points of previous titration curves.The solution contains salt and free acid, considered to be completelyionized. Therefore

1 fC\ 1\

- 7.63 X

pH = -log 7.63 X 10-4

pH = 3.12

Table 12 lists the calculated pH for 18 points for this titration. Theseare plotted in Fig. 22. By arguments similar to those given for other titra-

Page 134: quimica inorganica cuantitativa


10 15 20 25 30 35 40 45Ml. HCI

FIG. 22. Titration curve for NII 4OH with 0.1 TV HCI. Thirty ml. of 0.1 NNH 4OH diluted to 100 ml. before titrating.

Note the rather pronounced change of slope at about 5 ml. of HCI due to the

repression of the ionization of the weak base by the salt, NH 4C1, which by nowhas been formed in considerable quantity. From this point on to about 25 ml. of

HCI both weak base and salt of that base are present in fair amounts; as a result

the solution is buffered in this region. From 25 ml. to 30 ml. of acid the buffer

action again is very slight since only small amounts of base remain and the pHresumes its rapid fall with small increments of acid.

Page 135: quimica inorganica cuantitativa


tions and by inspection of this curve it should be clear that phenol-

phthalein cannot be used as the indicator in the titration of such a weak

base as ammonium hydroxide by hydrochloric acid, and that either

methyl red or methyl orange may be used, preferably the former.

Table 12


Titration of a Weak Base by a Weak Acid. Salts of weak acids and

weak bases hydrolyze to a considerable extent; therefore, in the titration

of a weak base and a weak acid hydrolysis plays an unusually important

role. The pH at the stoichiometric point depends upon the relative values

of Kb and Ka . If Kb and Ka are equal the titration results in a pH of 7.0

at the stoichiometric point. There will be practically no steep portion to

the titration curve (see Fig. 23), and thus no color indicator will reveal the

end point distinctly. Such titrations may be made potentiometrically but

reliable results cannot be obtained colorimetrically. Nevertheless muchcan be learned by constructing the titration curve. Suppose, as usual, we

consider 30 ml. of 0.1 N HC 2H 3 2 diluted to 100 ml. and titrated with 0.1

N NH 4OH.1. The calculation of the pH in the beginning, before any base has

been added, is, of course, identical with that already made for the acetic

acid-sodium hydroxide titration. The pH is 3.13,

Page 136: quimica inorganica cuantitativa


2. The solution corresponding to all other points preceding the stoichiometric

point consists of free acetic acid and the salt, ammonium acetate that is, a

weak acid and a salt of that acid. The solution therefore is a buffer. Accordingly,

the calculations of these points of the titration curve are made in the same man-ner as those for the titration of acetic acid with sodium hydroxide as computed by

equations (25) and (25a). The two curves are identical for corresponding additions

of base up to a volume of 29.0 ml.

At about 29.0 ml. of base, however, a difference in the two curves begins to

reveal itself. Anticipating the stoichiometric point for the present titration, it is

foreseen that the pH will be 7.0 since acetic acid and ammonium hydroxide have

ionization constants almost identical, 1.8 X 10~ 6. It follows then that the acetic

acid-ammonium hydroxide titration cannot possibly yield a solution with a

pH as great as 7.0 before 30 ml. of base has been added. In the case of the acetic

acid-sodium hydroxide titration after 29.90 ml. of base the pH was calculated as

7.22. Obviously then equation (25) will not serve for computing the pH for a

weak acid-weak base titration for points on the acid side but very close to the

stoichiometric point. Two questions arise then: why does equation (25) fail here,

and how may the calculation be made?

Recalling equation (25) it will be remembered that no account of hydrolysiswas taken. But the salt of a weak acid and a weak base hydrolyzes to such an

extent, even in the presence of a small quantity of acid or base, that this simplified

equation is no longer strictly valid. If a solution of such a salt contains also an

excess of weak acid, of concentration a, the total concentration of HA will be

a + y, where y is the concentration ofHA formed by hydrolysis of the salt; and

equation (25) must be replaced by the more exact expression


or, since near the stoichiometric point C is very great compared with y,

(25b) [H+]= ^-~- Ka

It is necessary to find the value of y before equation (25b) may be used. This maybe done by rearranging equation (17), namely,

[BOH][HA] Kv

taking into consideration that the acid concentration is greater than [HA] by y,

the concentration of HA due to hydrolysis. Then also [BOH] will be equal to yand since [B+] = [A"] = C y we have

(17a)y(a + y) K.

(C -y)' KaKk

or, since near the stoichiometric point C is great compared with y, we have the

excellent approximation8

8 y may be discarded in the denominator but not in the numerator because near

the stoichiometric point the concentration of the acid, a, is small, whereas that of the

salt, C, is large. Thus (C y) is practically identical with C, but (a + y) is by nomeans equal to a.

Page 137: quimica inorganica cuantitativa



In solving for y the above equation may be put into the form



from whence

a a

Once y is determined it may be inserted into equation (25b) and the value of [H+]

and then of pH may be obtained.

Suppose we calculate y in the titration of our acetic acid solution after the

addition of 29.90 ml. of ammonium hydroxide, that is, when 0.10 ml. of acid


a - (0-WWO.lOOO) = 7.70 X 10"

C = <-"" -1000) = 2.30 X 10-LZy.y

Substituting in equation (3 la),

7.70 X IP" 5

,(T^Tx IP" 5

)*" (2.30 X 10" 2)2(IP"


" 5

, /(T^Tx IP" 5

+ \ I* ~

2 \ I (L8 X 10- 6)(1.8 X 10- 6


= -3.85 X 10~ 5 + 13.35 X 10~ 6

y = 9.50 X 10- 5

And substituting this value of y in equation (25b),


]= 1.35 X 10-7

pH = 6.87

This compares with a pH value of 7.22 (see Table 10) which would be obtained

by using equation (25). In constructing the curve of Fig. 23 the value, 6.87, has

been used for the addition of 29.90 ml. of base. Equations (31a) and (25b) must

be employed if the degree of hydrolysis is 5 per cent or greater. The simpler equa-

tion (25) suffices for points more removed from the stoichiometric point than 0.1

ml. in such a titration as we have considered because hydrolysis is repressed when

considerable quantity of acid (or base) is present. To demonstrate this fact the

pH may be calculated by equations (31a) and (25b), and again by equation (25),

for the addition of 20 ml. of ammonium hydroxide. By either method a pH of

5.05 is obtained. Even with only 1 ml. of acetic acid remaining (after the addition

of 29 ml. of base), the pH calculated by either method is found to be 6.21.

3. At the stoichiometric point equation (18) gives the value of [H+]


(10-14)(1.8 X 1C

1.8 X 10- 8

[H+] = 10- 7

pH = 7.00

Page 138: quimica inorganica cuantitativa



FIG. 23. Titration curve for HC 211 O, with 0.1 N NH 4O1I. Thirty ml. of 0.1 NHCaliaO; diluted to 100 nil. before titrating.

4. When excess ammonium hydroxide has been added the solution contains

free weak base and its salt; therefore buffer equation (26) may be used exceptwhen very near the stoichiometric point where a more exact form must be used,that is


where b is the concentration of the excess base and y is the concentration of BOHdue to hydrolysis. The derivation is analogous to that of equation (25b). Thevalue of y is obtained in the same manner as y of equation (3la) and its formula-tion evidently becomes

(31b)b, t&,C*K.~2+\ I KJC*

To illustrate, when 30.10 ml. of ammonium hydroxide is added to the acetic acid

solution, equation (26b) leads to a pH value of 7.13 which is 0.13 unit on the basic

side of neutrality, the same extent by which 29.90 ml. of ammonium hydroxide

Page 139: quimica inorganica cuantitativa


failed to reach neutrality in this titration. The detailed calculation follows. Wehave present in the solution 0.10 ml. of excess base and 30 ml. of salt in a total

volume of 130.1 ml.

b = - = 7.69 X 10-

= , 31 x 10_

Therefore, __ _ 7.69 X IP" 5

/C7.69 X IP" 6)2

(2.31 X IP"2)2(10~


y ~2

+ \ * (1 8 X 10- 5)(1.8 X 10- 6


= -3.81 X 10- 5 + 13.39 X 10- 5

y = 9.55 X 10' 5

and substituting this value of y in equation (26b)

__ (10-")(2.3i X10-2)_


[(7.69 X TO- 5) + (9.55 X 10-"5)][1.8 X 10-']

[IP] = 7. It X 10- 8

pH = 7.13

The simplified equation (26) leads to the same pH values as does the more

exact calculation when the stoichiometric point has been passed to an extent of 1

ml. or more of ammonium hydroxide. This means of course that the hydrolysis

is repressed enough in the presence of greater amounts of base so that equation

(26) is sufficiently accurate.

The curve of Fig. 23 shows that there is no abrupt change in pHaround the stoichiomctric point. Thus il is impossible to find an indicator

which will detect the end point for a weak acid-weak base titration, and

such titrations are not attempted. It should be noted that the first part

of a titration curve of this type resembles that of a weak acid-strong base

from the beginning of the* curve until near the stoichiometric point. Fur-

thermore this curve resembles that of a strong acid-weak base titration

(if base is in the buret) in that part shortly after the stoichiometric point

has been passed and on to points corresponding to considerable excess

base. The question might arise then as to why we could not employ suc-

cessfully an indicator with a pll range from about 6.5 to 7.5 to mark the

end point of a weak acid-weak base titration. The answer is that in the

vicinity of the stoichiometric point this curve is not rising so abruptly (as

the exact calculations of pll for 29.90 and 30.10 ml. of base have shown)

as that resulting when either the acid or the base is strong. Thus an indi-

cator, even though giving a color change between pll 6.5 and 7,5, would

not be satisfactory since it would not give a sharp end point.

Other Titrations. Very Weak Acids, Polybasic Acids and Car-

bonates. The weaker the acid being titrated the higher will be the initial

pH. Thus the titration curves show less abrupt change of slope at the

stoichiometric point, and in order for such titrations to be feasible more

and more sensitive indicators are needed. Boric acid which has a value of

Page 140: quimica inorganica cuantitativa


Ka of 6 X 1Q- 10 cannot be titrated. See Fig. 21. However, if glycerol or

maimitol is added the boric acid is converted into complex acids which are

much stronger than boric acid itself. Under such conditions a good break

in the curve is obtained and phenolphthalein can be used as an indicator.

When polybasic acids are titrated with a strong base breaks may occur

corresponding to each step in the ionization of the acid. However, the

breaks in the curve appear as distinct steps only if the two or more ioniza-

tion constants differ widely in magnitude (KaJKa^ 5 10 4). Thus sulfuric

acid, though dibasic, acts like a monobasic acid since both steps in its

ionization are fairly complete. On the other hand, phosphoric acid gives

two rather distinct breaks in its titration curve, but not three. The first

break, Fig. 24, comes after the first hydrogen has reacted, the second

break when the second hydrogen has reacted, but there is no break indi-

cating the reaction of the third hydrogen. These facts are understood

when it is realized that Kai and Kfl2for phosphoric acid have values around

10~3 and 10~ 8respectively, and the third ionization constant, K<^ = 10~ 13


is near the value of Kw . A solution of NasP04 is strongly hydrolyzed,which is another way of saying that the trisodium salt will not be formed

by the reaction of the HP0 4" ion with sodium hydroxide. The curve in

Fig. 24 for phosphoric acid exhibits the first break at a pH near 4 and the

second at around 9. Thus methyl orange and phenolphthalein serve to

indicate these two stoichiometric points; the volume of base necessary for

the methyl orange end point is approximately one-third the quantity,and that necessary for the phenolphthalein end point is approximatelytwo-thirds the quantity, of base equivalent to all of the replaceable

hydrogen.A carbonate may be titrated by strong acids and two successive reac-

tions occur:

COr + H+ -> HC03-

HCOr + H+ - II 2C0 3

The second of the above reactions does not occur as long as any C0 3~

ion remains in the solution. When enough acid has been added to satisfy

the first reaction we find a break in the titration curve results. The further

addition of acid causes the formation of carbonic acid and at the comple-tion of the second reaction a second break in the titration curve is found.

The pH values at the two breaks are calculable. That established by the

bicarbonate ion has already been computed (p. 97) and found to be 8.4.

(If the titration is sodium carbonate against hydrochloric acid the solu-

tion will contain, when the first stoichiometric point is reached, sodiumbicarbonate and sodium chloride, but the latter will have no effect uponthe pH.) Therefore phenolphthalein may be used to detect this end point.

The pH of the solution corresponding to the second break in the curve is

Page 141: quimica inorganica cuantitativa





10 15 20 25

Ml. from buret

FIG. 24. Titration curves for H3P0 4 with 0.1 N NaOH (Curve 1) and for Na 2CO 3

with 0.1 N HC1 (Curve 2).

calculated by means of equation (6a). If the titration were 30 ml. of 0.1 A/

Na 2CO 3 diluted to 100 ml. against 0.1 N HC1 the concentration of car-

bonic acid would be


10-2)(3.0X 10-7


[H+] = 5.9 X 10-s

pH - 4.2

Page 142: quimica inorganica cuantitativa


This value comes within the range of methyl orange so that it may be

used as the indicator for the second step. Obviously carbonates may be

titrated first using phenolphthalein and then, after the red color has dis-

appeared upon the addition of acid, using methyl orange. The volume of

acid needed to reach the methyl orange end point will be just about double

the volume necessary to reach the phenolphthalein end point. Fig. 24

shows the curve for sodium carbonate titrated with hydrochloric acid.

The two end points in a carbonate titration make possible so-called

double titrations, in which carbonate and hydroxide, or carbonate and

bicarbonate, or hydroxide and bicarbonate,9 may be determined in the

presence of one another. Suppose a mixture of sodium carbonate and

sodium hydroxide is brought into solution and titrated with standard

hydrochloric acid. The following reactions will occur

(1) NaOH + HC1 -> NaCl + H 2

(2) Na2C0 3 + HC1 -> NaHC03 + NaCl

(3) NalICO, + 11C1 -> H 2C0 3 + NaCl

At the start of the titration the pll of the solution will be around 11 or

above and phenolphthalein will impart its red color to the solution. When

enough acid, say A ml., has been added to complete reaction (1) and (2),

the pH of the solution will be that due to sodium bicarbonate, or about

8.4, and the red color practically disappears. If now methyl orange is

added the color will be orange ; further addition of acid until the red color

of methyl orange appears, say 13 ml., marks the completion of reaction

(3). The A nil. of acid obviously is the volume necessary to react with

all the sodium hydroxide plus half enough for complete reaction with the

carbonate. The additional B rnl. of acid is the volume necessary to convert

the bicarbonate to carbonic acid, or 2B nil. is equivalent to all of the

sodium carbonate. Thus (A + B) 2B, or A B rnl. of acid are equiva-

lent to the sodium hydroxide. From these volumes, A B ml. and 2B

ml., and the normality of the acid, the amount of sodium hydroxide and

sodium carbonate in the sample may be calculated.

9 Mixtures of NaOH and NaHCO 3 are not encountered in solution because hy-

droxyl and bicarbonate ions react: OH~ -f HCO.r > COr + H2O. However, a drymixture of the two can be analyzed by the double-titration method since, even thoughwhen brought into aqueous solution the above reaction takes place, the stoiehiometric

relationship between the titrating acid and the ions actually present in the solution

will be the same as if the above reaction had not taken place. For example, suppose a

dry mixture contains one mole each of hydroxide and of bicarbonate. Obviously one

mole of a monobasic acid is equivalent to each of the two components or two moles to

the whole mixture. In solution, however, th hydroxide and bicarbonate react to form

a mole of sodium carbonate. To titrate on* mole of carbonate a mole of acid is neces-

sary to satisfy the titration to the plunolj'athalein end point, and an additional mole is

necessary to carry the titration (of the bicarbonate now formed from the carbonate) to

the methyl orange end point. Thus tb ) stoichiometric relationship is the same as it

would have been had no reaction taken place initially between OH~ and HCOr.

Page 143: quimica inorganica cuantitativa


The phenolphthalein end point in the above double titration is not

very sharp and a precision of better than 5 or 10 parts per 1000 can hardlybe obtained. Better results may be had if two aliquot portions of the

solution of mixed alkalies are used. The total of carbonate and hydroxideis first determined by titration against standard acid to the methyl orangeend point. Then, using the other portion, the carbonate is precipitated byadding a slight excess of barium chloride, and the hydroxide alone is

determined by titrating with acid to the phenolphthalein end point. The

precipitated barium carbonate need not be filtered off since at the phenol-

phthalein end point the pll is sufficiently high to prevent the barium

carbonate from dissolving.

Problem. A sample of a mixture of NaOH, Na 2C0 3 and inert material weighing0.2375 g. requires 34.06 ml. of 0.0960 TV HC1 to reach the phenolphthalein end

point and an additional 13.78 ml. to reach the methyl orange end point. What is

the per cent NaOH and Na 2CO 3 in the sample?Solution (a). Here A ml. = 34.06 and B ml. = 13.78.

Therefore 2B = 27.56 ml. acid equivalent to Na 2C0 3 .

A - B = 20.28 ml. acid equivalent to NaOH.TV . V = N' . V

(0.0960) (27.56) = N' 1000

TV' = 0.002646 = normality with respect to Na 2C0 3 , assuming the

sample is dissolved in 1 1. of solution.

N . V = N' . V(0.0960) (20.28) = N' 1000

N' = 0.001917 = normality with respect to NaOH, assuming the

sample is dissolved in 1 1. of solution.

(0.002646) (53.00) (100)


(0.001947) (40.01) (100)

59.0%Na 2C03

= 32.8% NaOH0.2375

Solution (b). The problem may be solved directly in terms of gram equivalent

weights in the following manner.The g. eq. wts. of Na 2COj present in the mixture must be equal to (0.02756)

(0.0960) since this is the g. eq. wt. of acid necessary to react with the carbonate.

The equivalent weight of Na 2C0 3 is 53.00. Therefore




In a like manner

(0.02028) (0.0960) (40.01) (100)


Use of Carbon Dioxide-Free Water. Since the atmosphere alwayscontains some carbon dioxide, solutions exposed to air will absorb the gasand form carbonic acid. In acid-base titrations the dissolved carbon diox-

ide will make no difference if, for example, methyl orange or methyl red is

used as the indicator. If, however, an indicator like phenolphthalein is

Page 144: quimica inorganica cuantitativa


used it is necessary to employ carbon dioxide-free water in the titration.

Consider an acid being titrated with sodium hydroxide and suppose that

the solution of the acid also contains carbon dioxide. When a quantity of

base chemically equivalent to the acid has been added the next excess drop

of sodium hydroxide should elevate the pH to a point where the phenol-

phthalein becomes red, thus marking the end of the titration. With car-

bonic acid present, however, the "excess" drop of sodium hydroxide will

react with the carbonic acid to form sodium bicarbonate. Further addi-

tion of base results in the formation of more bicarbonate until finally no

more carbonic acid remains. The pH at this point will be that estab-

lished by bicarbonates namely, about 8.4 so that the phcnolphthalein

is still colorless. The next drop of base will form sodium carbonate,

Na 2CO 3 ,and even in small concentration the presence of the normal car-

bonate will bring the pH up to the phenolphthalein range and, at last, the

red color appears. However, by this time an amount of base has been

added in excess of the amount equivalent to the acid being titrated by

just that quantity necessary to convert the carbonic acid into bicarbon-

ate. A considerable error thus has been introduced. Had methyl orange

been used as the indicator, the formation of the first of the bicarbonate

would have caused the pH to be on the orange side of the indicator and no

harm would have resulted. Since indicators with a high pll range must be

used in the titration of weak acids by strong bases it follows that carbon

dioxide-free water 10 must be employed in such cases. In general one

should use carbon dioxide-free water if the indicator employed changes

color at a pH above 8.4, i.e., that established by bicarbonates.


It is worth repeating that for the calculation of the pll of the solution

being titrated there are, in general, four typical calculations to be made.

These are: (1) the pll of the solution before the addition of any reagent

from the buret; (2) the pH after the addition of reagent from the buret

but before the stoichiometric point has been reached; (3) the pH at the

stoichiometric point; (4) the pH after excess reagent has been added. The

method of making the above four kinds of calculations has been explained

(a) for the titration of a strong acid with a strong base, (b) for the titra-

tion of a weak acid with a strong base, (c) for the titration of a weak base

with a strong acid and (d) for the titration of a weak acid with a weak


In addition to the four typical calculations listed above, in which it is

sufficient to employ equations which are sometimes close approximations

10 Prepare carbon dioxide-free water as follows: Place about 4.5 1. of distilled water

in a 5 I pyrex flask and boil to expel C0 2 . Insert a two-hole rubber stopper equippedwith a siphon and a soda-lime tube.

Page 145: quimica inorganica cuantitativa


only, there are a few instances in which approximate equations are not

valid enough to reveal the exceptional nature of the calculation involved.

The exceptions are as follows. (1) For the calculation of [H+] in the titra-

tion of a weak acid by any base, in the early stages of the titration, in-

stead of equation (25) one must use (25a). (2) Likewise, for calculation of

[OH~] (and thence [H+]) of the solution in the titration of a weak base

with any acid, in the early stages of the titration, instead of equation (26)

one must use (26a). (3) In the titration of a weak acid by a weak base

(not practicable, of course, with color indicators), in addition to excep-tions (1) and (2) just cited, the calculation of [H+] established just before

and just after the stoichiometric point must be made by using, instead of

equations (25) and (26) respectively, equations (25b) and (26b).

The conditions under which the two buffer equations, (25) and (26;,

do not lead to good approximations of the hydrogen ion concentration,

and the more exact equations which, under these conditions, replace

them, are brought together below.

Equation (25) [H+] ^

Equation (25a) X =[I

p * Acid buffer; no complications.

Equation (25b)

Use instead of (25) for exact [H+] for points in early

stages of weak acid-base titration.

[H+]= L+J - Ka

Use instead of (25) for exact [H+] for points near

stoichiometric point (on acid side) in weak acid -weak

base titration. (Obtain y from (31a).)

K C[H+] = nrrktiTfc^' Basic buffer; no complications.Equation (26)

Equation (26a) X = [OHh] = -

+ [BOH]*,

Equation (26b) [H+]

Use instead of (26) for exact IOH~] (whence exact [H+])

for points in- early stages of weak base-acid titration.


Use instead of (26) for exact [H+] for points near stoi-

chiometric point (on basic side) in weak acid-weak base

titration. (Obtain y from (31b).)

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a /a* C*KEquation (31a) Y = ~

2+ V 4

+ rfUse in connection with (25b).

T- , /oi^ ,

Equation (31b) y = + / +

Use in connection with (26b).

Questions and Problems

(See Appendix for ionizulion constants)

1. Calculate the pi I established by the following solutions:

(a) A 0.02 /V solution of sodium beiizoate.

(b) A 0.1 /V solution of ammonium fluoride.

(c) A solution of potassium bitartratc ( f moderate concentration.

(d) A solution which is 0.123 TV with respect to benzole acid and 0.246 TV with

respect to sodium benzoate.

(e) One hundred ml. of a solution identical with that of (d) except that it also

contains 10 ml. of 0.1 TV hydrochloric acid.

Answer to (d): 4.5.

2. The ionization constant of a weak acid, HA, is 1.86 X 10~ 5. Calculate the pH

of a 0.1 /V solution.

Answer: 2.87.

3. The ioni/ation constant of a weak base, BOH, is 3 X 10~ 6. Calculate the pll

of a 0.1 N solution.

4. (a) Twenty-five ml. of a 0.1 /V solution of a weak acid (K = 10~ 6) is diluted

to 100 ml. and titrated with 0.1 N potassium hydroxide solution. Calculate

the approximate pH of the solution resulting from the titration at a sufficient

number of points to reveal the path of the titration curve. Plot these points

on graph paper, (b) Repeat (a) using 0.01 N solutions of both acid and base.

(c) What indicator would you select for the titration of (a)? Would the

same indicator be suitable for the titration of (b) ?

5. Derive equation (25a) from equation (25) after inserting into the latter the

more exact expressions which take into account the slight ionization of the

weak acid. (Cf. p. 111.)

6. (a) In the titration of 4 (a) above what is the difference in the value of the pllas calculated by equation 25 and by equation (25a) after the addition of 1.00

ml. of base? (b) How does this difference, A pi I, vary for a given small addi-

tion of base to weak acids having different values for Ka $

Answer to (a): A pH = 0.10.

7. When distilled water is exposed at room temperature to the atmosphere it will

absorb carbon dioxide to the extent of about 0.000014 mole per liter. Calcu-

late the pll of the water.

Answer: 5.7.

8. Explain the type and extent of the error introduced in titrating a weak acid

with sodium hydroxide solution when the solution also contains dissolved

carbon dioxide.

Page 147: quimica inorganica cuantitativa


9. A solution is made up by diluting exactly 5 ml. of 0.1 N hydrochloric acid to

100 ml. (a) Calculate the pH of the solution, (b) How many grams of solid

sodium acetate, NaC 2H 3 2 , must be added to this solution (assume that the

volume remains constant at 100 ml.) in order to change the pH to a value of


Answer: (a) 2.3; (b) 0.0745 g.

10. (a) A solution is prepared by diluting 2.5 ml. of 0.2 TV sodium hydroxide to

100 ml. Calculate the pll of the solution, (b) How many grams of solid

ammonium acetate must be added to this solution in order to change the pHto a value of 9.34?

11. A 0.025 M solution of the sodium salt of a certain weak acid has a pH of 8.57.

Calculate the value of Ka for the arid.

Answer: 1.8 X 10~ 5.

12. A solution of the ammonium salt of a certain weak acid has a pH of 6.75.

Calculate the value of Ka for the acid.

Page 148: quimica inorganica cuantitativa

Chapter 8


IN DETERMINING the acid or base present in a sample unknown in respect

to one of these constituents it is only necessary to learn what quantity of a

standard solution of a base or an acid is required to react with the un-

known. Thus it is customary to have on hand a standard solution of an

acid and a base which may be used in such determinations. The acid most

commonly employed is hydrochloric although sometimes sulfuric and

oxalic acids are used. The last two are not volatile and accordingly may be

boiled, if desired, with the unknown without loss. The base most com-

monly used as a standard solution is sodium hydroxide but potassium

hydroxide and barium hydroxide often are chosen. The standardization

of the solution, acid or base, simply involves the titration of the solution

against an accurately weighed quantity of some basic or acidic substance

which meets the criteria already listed on p. 78 for primary standards.

Of course, once a solution of either has been standardized, it may be

titrated against the solution of the other and the normality of the latter

thus determined. In this case, however, the concentration of the latter has

been ascertained through a secondary standardization and the accuracyof the normality is subject to greater error than when a primary stand-

ardization has been carried out. In the following sections directions will

be given for the standardization of a hydrochloric acid solution and for a

sodium hydroxide solution, both about 0.1 TV; also the procedures for

certain analyses in which these two standard solutions are used. It should

be obvious that standard solutions which are about 0.1 AT will yield moreaccurate results in analyses than, say, 0.5 N solutions since a given weightof an unknown will require live times the volume in the former case. Theamount of the sample taken for analysis and the concentration of the

standard solution used should be such that, as a rule, a titrating volumeof from 30 to 50 ml. will result. Too small a titrating volume may intro-

duce an unnecessarily large error, and volumes over 50 ml. require refilling

a 50 ml. buret which also increases the chance of error.


The object is to prepare a solution of hydrochloric acid which will be

in the neighborhood of 0.1 TV, and then to learn through its standardiza-

tion exactly (with a precision of about 1 part per 1000 possible deviation)130

Page 149: quimica inorganica cuantitativa


what the normality is. The equivalent weight of hydrochloric acid is

approximately 36.5, so that the solution which is prepared must contain

around 3.65 g. of hydrogen chloride per liter of solution. From Table 6,

p. 84, we learn that 1 ml. of concentrated hydrochloric acid contains

0.43 g. of HC1. Thus to make 1 1. of 0.1 TV solution we shall need about

8.5 ml. of the concentrated acid.

Procedure. With a graduated cylinder, measure 9 ml. of concentrated

hydrochloric acid and transfer to a 1 1. glass-stoppered bottle. Add about

1 1. of carbon dioxide-free water and shake thoroughly. Keep the bottle

stoppered at all times. After filling a buret with the solution always re-

place the stopper in the bottle at once since evaporation would cause a

change in the normality. If desired, the bottle may be equipped with a

siphon for delivery ;in that case use a two-hole rubber stopper and insert a

soda-lime tube to prevent the entry of carbon dioxide and moisture from

the air.


Standard solutions of sodium hydroxide cannot be prepared directly

i.e., by weighing the hydroxide accurately, dissolving and diluting to

proper volume because it is hygroscopic and because it always contains

some sodium carbonate. In some analyses it is necessary to use carbonate-

free sodium hydroxide solution; it therefore is well to prepare a solution

containing little or no carbonate. This may be done by adding to an

ordinary solution of sodium hydroxide which is contaminated with some

carbonate, sufficient barium chloride to precipitate the carbonate. Thesolution is then filtered through asbestos in a Gooch crucible. This methodof course calls for a slight excess of barium chloride; if this is objectionable

the method given in the procedure below is recommended.

Procedure. Into a warmed clean bottle of 1000 to 1200 ml. capacity

pour about 50 ml. of melted paraffin. Turn the bottle on its side and roll

so that as the wax solidifies it coats the inside of the bottle. While some of

the paraffin is still liquid turn the bottle upright so that the bottom re-

ceives a coating of wax. 1 When cool pour in a liter of freshly boiled and

cooled, distilled water (carbon dioxide-free water).

Prepare a Gooch crucible with an asbestos mat as previously described

on p. 10.

The equivalent weight of sodium hydroxide is about 40; therefore for

1 1. of 0.1 N solution, 4 g. is needed. However, because of the superficial

coating of sodium carbonate on the hydroxide and the hygroscopicity of

the base, 4.5 g. of the highest grade sticks or pellets of sodium hydroxide

1 Instead of paraffin wax one may use Thermoprene, No. 1023, a rubber paint sold

by B. F. Goodrich Co., Akron, Ohio. See Soule, Ind. Eng. Chem., Anal. Ed., 1, 109


Page 150: quimica inorganica cuantitativa


should be weighed out using a rough balance. Furthermore, some sodium

hydroxide will be lost during the filtration called for below. After weigh-

ing, transfer to a small, glass-stoppered Pyrex bottle or test tube, add

about 5 ml. of water and shake until solution is complete. A cloudiness will

remain; this is the sodium carbonate which precipitates out in concen-

trated sodium hydroxide solution. After the solution is cool filter it

through the Gooch crucible, pour the clear filtrate into the liter of water

in the paraffin-coated bottle and shake. If desired the bottle may be

equipped with a siphon for delivery; in that case use a two-hole rubber

stopper and insert a soda-lime tube to prevent entry of carbon dioxide

from the air.

Relative Concentrations of the Acid and Base. Fill two burets,

one with the hydrochloric acid solution and the other with the sodium

hydroxide solution. Be sure to rinse each buret with three successive 8 or

10 ml. portions of the solution with which it is to be filled, transferring

the solution directly from the storage bottle to the buret.

Run 30.00 ml. of the acid from the buret into a 250 ml. beaker or into

a 125 ml. Erlenmeyer flask. Dilute with about 50 ml. of carbon dioxide-

free water. Add a drop or two of phenolphthalein indicator. Place the

beaker under the buret filled with sodium hydroxide solution and titrate

until a faint pink color is obtained which persists for 30 seconds. (After

reading the buret prove that carbon dioxid ecauses the pink color to fade

by blowing the breath through a glass tube into the pink solution.) Fromthe corrected volumes of acid and base calculate the ratio of the volume of

acid to base. The deviation from the mean should not exceed 1 part per1000.

Notes. Read carefully p. 125 on the use of carbon dioxide-free water. If the

titration is carried out in a beaker, instead of in an Erlenmeyer flask, stirring

should be done with a gentle, circular motion. Whipping the solution with a

stirring rod will introduce carbon dioxide from the air.

Repeat the above experiment using either methyl orange or modified

methyl orange indicator. 2

Record the results in ink in the notebook, using some such form as

suggested in Table 13.


The primary standard which is used to standardize hydrochloric acid


is sodium carbonate using, of course, methyl orange or

2Prepare modified methyl orange: 1 g. methyl orange and 1.4 g. xylene cyanole FF

in 1 1. of water, or purchase already prepared from Eastman Kodak Co . (See p. 104.)3 Before performing this or any other laboratory experiment of this chapter, the

student should review thoroughly all topics taken up in Chapter 7.4Hydrochloric acid can be standardized gravimetrically by precipitating the

chloride ion and weighing as AgCl. See p. 248.

Page 151: quimica inorganica cuantitativa


Table 13


(Phenolphthalein Indicator)

* The true volume is the burct reading plus the buret correction for the particular volume; see p. 71.

any other indicator changing color at a pll around a value of 4. If sodium

carbonate of a reliable grade is not available it must be prepared as de-

scribed here.

Preparation of Sodium Carbonate. Place about 5 g. of pure sodium

bicarbonate in a porcelain or a platinum crucible and imbed the crucible to

three-fourths its height in a sand bath. Insert a 360 thermometer in the

sand bath alongside the crucible and heat until the temperature registers

between 275 and 290 for 1 hour. At this temperature the bicarbonate

undergoes the reaction

2NaIIC0 3- Na2C0 3 + H 2 + C0 2

Do not allow the temperature to exceed 300 at the most, since above that

temperature sodium carbonate decomposes:

Na 2C0 3 -> Na 2 + CO 2

and the final product would be a mixture of sodium carbonate and sodium

oxide. Cool and weigh the crucible and contents, after which heat again

for 20 minutes, cool and reweigh. The process of heating and weighing is

continued until a constant weight is obtained. Finally transfer the puresodium carbonate to a glass-stoppered bottle while warm and place in a


Procedure for Standardization. It is evident (see p. 79) that, since

the equivalent weight of sodium carbonate is 53, the quantity suitable for

Page 152: quimica inorganica cuantitativa


titrating with tenth-normal acid is from 0.16 to 0.22 g. Accurately weigh

three portions of about this amount of the sodium carbonate and transfer

(see p. 80) to 250 ml. beakers or to Erlenmeyer flasks. Dissolve each in

about 100 ml. of water, add 2 drops of methyl orange (or modified methyl

orange) and titrate with the hydrochloric acid. With methyl orange indi-

cator the titration is carried to a point where the yellow solution assumes

a very faint orange color. Wash down the walls of the beaker with water

from the wash bottle; if the color becomes yellow again add acid from the

buret dropwise until the orange color is restored. (If the end point is

accidentally overrun i.e., if the color of the solution becomes distinctly

pink countertitrate with the sodium hydroxide solution until the faint

orange color is restored. Since the relation of the acid and the base solu-

tions has been determined, the appropriate correction for the volume of

acid equivalent to the sodium carbonate may be computed.) From the

weights of the sodium carbonate and the volumes of hydrochloric acid

employed the number of grams of sodium carbonate equivalent to I 1. of

the acid may be calculated to four significant figures. These three results

should show an average deviation not greater than 1 part per 1000. Calcu-

late the normality of the hydrochloric acid using the method given on p.

81, equation (1). Record the data on some such form as that suggested


* True volume =* Buret reading plus huret correction (see p. 71).

Page 153: quimica inorganica cuantitativa


in Table 14. From the ratio of the concentrations of the hydrochloric acid

and the sodium hydroxide already determined calculate also the normal-

ity of the base. (Which ratio will you employ, that obtained using phenol-

phthalein as the indicator, or the one using methyl orange?)

Notes. Review methods of weighing, p. 5 Iff., and see p. 80

Preparation of Standard Hydrochloric Acid from Constant-

Boiling Acid. At a definite atmospheric pressure and a definite rate of

distillation the distillate from constant-boiling hydrochloric acid has a

definite composition.6 Table 15 gives the composition of the distillate ob-

tained at ordinary pressures when the procedure which is described be-

low is followed.

Table 15


Procedure. Place in a 1 1. distillation flask about 750 ml. of hydrochloric

acid of a density of about 1.18. Arrange a condenser in the usual manner

for distillations. Distil at a rate of 3 to 4 ml. per minute until about three-

fourths of the original liquid has been distilled. Reject this distillate and

continue the distillation, now collecting the constant-boiling liquid in a

new, dry flask. Discontinue the process when the residual liquid has

diminished to 50 or 60 ml. Record the barometric pressure to the nearest

millimeter during the collection of the sample.

To prepare a solution of hydrochloric acid which will be approximately0.1 TV from the constant-boiling solution, proceed as follows. Place about

18 ml. of the constant-boiling solution in a previously weighed glass-

stoppered weighing bottle. Weigh the bottle plus acid and quantitatively

transfer (see p. 9) the acid to a 1 1. volumetric flask, dilute to the markwith water within a degree or two of the temperature for which the flask

was calibrated, stopper and thoroughly shake. Calculate the exact nor-

mality of the solution.

5 Foulk and Hollingsworth, J. Am. Chem. Soc., 45, 1220 (1923).

Page 154: quimica inorganica cuantitativa


Problem. Constant-boiling hydrochloric acid, which was collected at a pres-

sure of 745 mm., weighing 17.9962 g. is diluted to 1000.0 ml. in a volumetric flask.

Calculate the normality of the solution.

Solution. N = ~ = 0.10005


As mentioned under the standardization of hydrochloric acid, the

normality of the sodium hydroxide solution may be obtained from the

normality of the former and the ratio of the concentrations of the two.

Often, however, it is desired that the normality of the base be established

through a primary, rather than a secondary, standardization. Primary

standards for solutions of strong bases are plentiful in number but the

three most commonly employed are potassium biphtlialate, sulfamic acid

and benzoic acid.

Potassium biphtlialate, C 6H4(COOH)COOK, has several advantages

to recommend it: it does not absorb moisture readily, so that it is easily

dried at 105 and can be accurately weighed, even in rainy weather; it can

be obtained in very pure form; it has a high equivalent weight and it is

soluble in water. The reaction

HC 8H4Or + OH- -> C 8H 4Or + H2

shows that the equivalent weight is the same as the molecular weight.

Sulfamic acid is a strong monobasic acid, nonhygroscopic, and it maybe obtained in a pure state for standardizations. 6 It too dissolves freely in

water with which it slowly reacts:

HS03NH2 + H2 -+ NH 4HSO4

However, the above reaction does not take place rapidly enough to inter-

fere with the standardization.

Benzoic acid, C 6H 5COOH, may be obtained from the U.S. Bureau of

Standards in 99.9 per cent purity. If an analytical grade of the acid is

purchased it should be carefully fused in a hot air oven at a temperature

below 125; at higher temperatures charring is likely to take place. After

the fusion the cake is crushed with mortar and pestle and ground to a

powder. In this form it can be accurately weighed more easily than the

fluffy form which it has before fusing. Since benzoic acid is so sparingly

soluble in water it must be dissolved in ethyl alcohol before titrating. Be-

cause the alcohol may be slightly acidic it is necessary to run a blank when

benzoic acid is used as a standard.

Procedure Using Potassium Biphthalate or Sulfamic Acid. Ac-

curately weigh out pure, dry samples of from 0.6 to 0.8 g. of potassium

6.G. F, Smith Chemical Co., Columbus, Ohio.

Page 155: quimica inorganica cuantitativa


biphthalate or 0.3 to 0.4 g. of sulfamic acid. (Confirm the fact that these

weights are suitable for titrating with 0.1 AT base; see p. 79.) Transfer the

samples to 250 ml. beakers or to Erlenmeyer flasks and dissolve in about100 ml. of carbon dioxide-free water. Add the indicator, 2 drops of

phenolphthalein if the biphthalate is the standard, 2 drops of phenol-

phthalein or any other indicator with a pH range occurring between 4and 9 in the case of sulfamic acid. Titrate until finally 1 drop of base fromthe buret brings about the color change, a faint pink persisting for 30

seconds if phenolphthalein is used. Record the data in a manner similar

to that suggested in Table 14 and calculate the normality of the sodium

hydroxide. The deviation from the mean should be about 1 part per 1000

or less.

Notes. Read carefully p. 125 on the use of carbon dioxide-free water. Remem-ber too that if the titration is carried out in a beaker, instead of in an Erlenmeyerflask, stirring should be done with a gentle, circular motion. Whipping the solu-

tion with a stirring rod will introduce carbon dioxide from the air. Review methodsof weighing, p. 5 Iff., and see p. 80.

Procedure Using Benzole Acid. Accurately weigh pure, dry samplesof benzoic acid weighing about 0.4 or 0.5 g. (see p. 79). Place the watch

glass containing the acid in a beaker and add 20 ml. of ethyl alcohol.

Swirl until the acid completely dissolves; this may be hastened by warm-

ing a minute or two over a steam bath. Be sure that all of the benzoic acid

is in solution; then add 50 ml. of carbon dioxide-free water and 2 drops of

phenolphthalein. Titrate with the sodium hydroxide solution until the

appearance of a faint pink color which persists for 30 seconds. A ring of

benzoic acid may have been deposited on the walls of the beaker justabove the level of the solution. Rotate the beaker in a slightly tilted posi-

tion so as to redissolve any such benzoic acid. If the pink color fades as a

result of this, add more sodium hydroxide dropwise until a faint pinkwhich lasts at least 30 seconds is restored.

Run a blank by combining 20 ml. of alcohol and 50 ml. of carbon di-

oxide-free water and adding 2 drops of indicator. Subtract any volume of

sodium hydroxide solution required to impart a pink color to the blankfrom the titration volumes in the standardization. From the net volumeof hydroxide and the weight of standard calculate the normality of the

sodium hydroxide solution. Results should not show a deviation from the

mean greater than 1 part per 1000.

Notes. Read notes following the procedure for standardization with potassiumbiphthalate.

A blank must be run because ethyl alcohol often contains small amounts of

acidic constituents which require some base for neutralization. Without the blank

determination the calculated normality would be too low.

Page 156: quimica inorganica cuantitativa




The minimum acid content of vinegar is set by law. A good qualityof vinegar will run about 5 per cent acetic acid by weight. Since the

density of vinegar is close to unity this means that there is around 5 g.

of acetic acid per 100 ml. of vinegar, or 50 g. per liter. Therefore vinegarsare approximately

5%o normal, since the equivalent weight of acetic acid

is about 60. It follows then that if the sample is titrated with 0.1 TV sodi-

um hydroxide the volume of base used will be roughly eight times that of

the vinegar. Accordingly a suitable amount of vinegar will be about 5

ml. But if a volume as small as 5 ml. is measured with a pipet the relative

error introduced will be quite large. For this reason it is necessary, if high

accuracy is to be attained, to weigh the sample of vinegar. If a sample of

about 5 g. is weighed only to the third decimal place a precision of 1 in

5000 is secured and this is better than the precision gained in reading the

volume of sodium hydroxide from the buret.

Procedure. Weigh a clean, dry glass-stoppered weighing bottle to the

nearest milligram. Using a dry 5 ml. pipet, or one that has been rinsed

three times with the vinegar to be analyzed,transfer about 5 ml. of vinegarto the weighing bottle. Weigh the bottle plus vinegar, again to the nearest

milligram. Transfer the sample quantitatively (see p. 9) to a 125 ml.

Erlenmeyer flask, add 50 ml. of carbon dioxide-free water and 2 dropsof phenolphthalein and titrate with standard sodium hydroxide solution

until a faint pink color appears which persists for 30 seconds. Calculate

the percentage by weight of acetic acid. The deviation should be under2 parts per 1000 for duplicate runs.

Notes. Further notes are unnecessary provided those at the conclusion of the

procedure for standardizations given in the preceding sections have been studied


Problem. A sample of vinegar weighing 5.082 g. is titrated with 0.1012 TV

sodium hydroxide, 42.82 ml. being required. Calculate the percentage by weightof acetic acid.

Solution. G. eq. wts. of base used = (0.01282) (0.1012). G. eq. wts. ofHC 2H 3 2

present must be same. The equivalent weight of HC 2H3 2= 60.05. Therefore

o/ HP H O - (0.04282)(0.1Q12)(60.05)(100)% ilCjzllaUjj = = 5.12D.UoZ


The determination of the acidity of a sample of oxalic acid typifiesthe analysis for total acid in a great many substances. Of course one al-

ways must bear in mind that the acid strength of the unknown determinesthe choice of the indicator and whether or not carbon dioxide-free water

Page 157: quimica inorganica cuantitativa


must be used. Therefore the principles discussed in Chapter 7 should be

recalled before embarking upon any acid-base titration.

Procedure. Accurately weigh about 1 g. of the sample, dissolve in a

small amount of carbon dioxide-free water in a beaker and quantitativelytransfer (see p. 9) to a 250 ml. volumetric flask. Dilute to the mark,

stopper the flask and shake. With a 50 ml. pipet withdraw three portionsand place in 125 ml. Erlenmcyer flasks. Add 2 drops of phenolphthaleinand titrate with standard sodium hydroxide solution. Calculate the per-

centage of acid present in terms of H^C^O^H^O. The deviation from the

mean for duplicate determinations should be 1 part per 1000 or less.

Notes. Instead of using aliquot portions from the single sample dissolved in

the 250 ml. of solution, one may, of course, weigh out two or more samples of

about 0.2 g. each. According to the above procedure any error made in weighingor dissolving and diluting to volume will obviously be proportionately transmitted

to each of the aliquot portions.


It has been mentioned already (see p. 121 and Fig. 21) that boric acid

is a very weak acid, having an ionization constant, Ka , of 6 X 10~ l. It

therefore cannot be titrated with a 0.1 TV solution of sodium hydroxide,for there will be no sharp break in the titration curve, which is to say no

indicator will clearly mark the end point of the titration. However, the

addition of polyhydric alcohols such as glycerol, mannitol or invert sugartransforms boric acid into comparatively strong complex acids which

make possible the titration with 0.1 TV alkali using phenolphthalein as the

indicator. Boric acid, H 3B0 3 , has a molecular weight of 61.84; it reacts

like a monobasic acid so that in calculating the acidity of a sample the

equivalent weight is also 61.84.

Procedure. Secure a sample which may be fairly pure boric acid or a

mixture of boric acid and inert material. If the sample were pure boric,

acid a 0.2 g. sample evidently would be a suitable quantity, but unless the

approximate composition is known it is best to make a preliminary deter-

mination by weighing out roughly 0.4 g. and, ignoring the refinements of

the titration, making a quick determination of the acid strength. If the

result were a titration, say, of 75 ml. it is obvious that a 0.2 g. sampleshould be taken for careful analysis; if the result were around 10 or 15

ml. a new sample of about 1 g. would be indicated. After accurately weigh-

ing the samples, transfer to Erlenmeyer flasks and dissolve in about 50

mi. of carbon dioxide-free water. For each 10 ml. of 0.1 TV boric acid

present (as indicated by the preliminary titration) add 3 or 4 ml. of 50 percent invert sugar, or 0.5 to 1 g. of mannitol, or 4 or 5 ml. of glycerol;

add also 10 drops of phenolphthalein. Titrate with standard sodium hy-droxide to a faint pink color. Then add another portion of 4 or 5 ml. of

Page 158: quimica inorganica cuantitativa


invert sugar, or the equivalent of mannitol or glycerol as indicated by the

above figures. The pink color probably will disappear; in this case slowlyadd sufficient base to restore a faint pink. Then add another 5 ml. of

invert sugar (or the equivalent in mannitol or glycerol). Repeat the alter-

nation of base and invert sugar until the color no longer fades. Calculate

the percentage of boric acid in the sample. A deviation from the mean of

duplicate determinations may be as high as 5 parts per 1000.

Notes. Suppose the preliminary determination required, for example, 12 ml.

of 0.1 N base. New samples for accurate analysis some three times as large wouldbe in order. This would mean that these samples contained the equivalent of 36ml. of 0.1 N boric acid; therefore approximately 10 or 15 ml. of invert sugarshould be added at the beginning of the titration. Be sure that the polyhydricalcohol which is used is neutral.

For reasons which should be obvious from previous titrations, carbon dioxide,if present, would cause high results. Therefore, the solution should be swirled

instead of shaken as the titration proceeds.



Soda ash is crude sodium carbonate and usually contains some im-

purities such as hydroxide, chloride, etc. Any hydroxide present con-

tributes, of course, to the total alkalinity of the soda ash. Analysis of soda

ash by titrating with standard hydrochloric acid solution therefore does

not give the percentage of Na 2C0 3 but the total basic strength of the

sample. The procedure is essentially the same as given for the standardi-

zation of hydrochloric acid using sodium carbonate as the standard. How-ever, instead of weighing out samples of 0.18 to 0.2 g. it is better to start

with a larger sample, dissolve and dilute to the mark in a volumetric flask

and pipet out aliquot portions which will contain around 0.2 g. each. In

this manner the inaccuracy resulting from the use of a small original sam-

ple and due to the heterogeneity of technical soda ash is largely avoided.

Procedure. Accurately weigh out two samples of about 1 g. each.

With a sample this great a precision of 1 part per 1000 is possible by weigh-

ing only to the third decimal place. Dissolve in a small amount of water in

a beaker and quantitatively transfer (see p. 9) to 250 ml. volumetric

flasks. Dilute to the mark and shake. Secure four portions for titration bytransferring, with a 50 ml. pipet, two aliquots from each of the two volu-

metric flasks to beakers or Erlenmeyer flasks. Dilute to about 100 ml.

and Citrate with standard hydrochloric acid using methyl orange or

mortified methyl orange as the indicator. Calculate the results in terms of

percentage of sodium carbonate or of sodium oxide. The deviation fromthe mean for titrations of two portions from the same volumetric flask

Page 159: quimica inorganica cuantitativa


should not exceed 1 part per 1000; that for all four titrations may run as

high as 2 parts per 1000.

Notes. The determination of alkali present in the soda ash as hydroxide maybe determined by adding barium chloride in excess to an aliquot portion of the

original solution. This precipitates the carbonate as barium carbonate. The pre-

cipitate need not be filtered off (see p. 125). Titrate with standard acid and calcu-

late the percentage of hydroxide.



It has been pointed out already that in the titration of carbonates the

first reaction with acid is that representing the conversion of the car-

bonate ion into the bicarbonate ion. After the completion of this reaction

the further addition of acid converts the bicarbonate ion into carbon diox-

ide and water. The end points for these reactions are recognized by using

two indicators, phenolphthalein and methyl orange. Reread pp. 124-125

for the full explanation of the principle of double titration, and then

analyze a mixture of carbonate and hydroxide as given in the following


Procedure (A). Weigh out a sample of about 2 g. of the mixture of

carbonate and hydroxide to the nearest milligram and dissolve in a small

amount of water in a beaker. Then transfer quantitatively (see p. 9) to

a 250 ml. volumetric flask, dilute with carbon dioxide-free water to the

mark and shake thoroughly. The solution, when titrated to the phenol-

phthalein end point, will contain sodium hydrogen carbonate which estab-

lishes a pH of about 8.4 (see p. 97). To recognize this end point prepare a

color comparator by making a solution of about 0.2 g. of sodium bicar-

bonate in 50 ml. of water and adding 5 drops of phenolphthalein. Keep in a

stoppered 125 ml. Erlenmeyer flask.

Now withdraw from the volumetric flask with a 25 ml. pipet two por-tions of the sample and place in 125 ml. Erlenmeyer flasks equipped with

stoppers. Add 5 drops of phenolphthalein. Remove the stopper only

long enough to add the standard hydrochloric acid during the titration.

Titrate until the faint pink color matches that of the bicarbonate color

comparator already prepared. Approach the end point by 4-drop incre-

ments of acid from the buret. Shake well after each addition of acid for a

full minute before making color comparisons.7

When the color match has been obtained read the buret; then add 2

drops of methyl orange indicator and continue titrating with standard

hydrochloric acid until the color changes from yellow to orange or to the

faintest pink. The flask need not be stoppered during this second part of

7Simpson, Ind. Eng. Chem., 16, 709 (1924), suggests the use of a mixed indicator to

detect this end point.

Page 160: quimica inorganica cuantitativa


the titration. Read the buret again. From the two readings calculate the

percentage of sodium hydroxide and sodium carbonate in the sample.

(See p. 125 for illustration.) Because the phenolphthalein end point is not

sharp the precision will not be high when the above procedure is followed.

The deviation for duplicate analyses may run as high as 10 parts per 1000.

Better results may be obtained by method (B) below.

Notes. A sample of approximately 2 g. dissolved in 250 ml. of solution would

yield about 0.2 g. in a 25 ml. aliquot portion. This obviously would require

roughly 35 to 50 ml. of 0. 1 N hydrochloric acid for a sample consisting of carbon-

ate and hydroxide. If, however, considerable inert material were present, a larger

original sample would be required to assure a titration involving over 30 ml. of

acid. If the procedure given above results in an inordinately low volume of acid

in the titration, it may be necessary to start a^ain with a larger sample.

In titrating to the phenolphthalein end point the flask should be kept stop-

pered while shaking not only to exclude atmospheric carbon dioxide, but to avoid

loss of carbon dioxide coming from the reactants. Local excess of hydrochloric

acid will cause the reaction, HC0 3

- + H H -> 11>O + CO 2 , to take place before

the desired reaction, C0.r + H+ -+ HC(V, is completed. This would cause loss

of carbon dioxide in an unstoppered ilask and give high results for the phenol-

phthalein end point. Shaking in the stoppered flask will cause the carbon dioxide

to redissolve and form the bicarbonate ion.

Procedure (B). Withdraw two 25 ml. aliquot portions from the volu-

metric flask containing the sample as secured by procedure (A) above.

With one portion determine the total alkalinity according to the pro-

cedure for the standardization of hydrochloric acid on p. 133, using

methyl orange. To determine the sodium hydroxide alkalinity add to the

other portion an excess (probably 10 ml. or less) of 0.5 M barium chloride

and titrate without filtering with standard hydrochloric acid using

3 drops of phenolphthalein as the indicator. Make up a blank con-

sisting of about 0.1 g. of sodium carbonate in 50 ml. of water and the

same volume of 0.5 M barium chloride solution as was used in the ti-

tration. Add 3 drops of phenolphthalein. The blank will probably be

white at this point. Add dropwise a standard solution of sodium hydrox-

ide from a buret until the first faint pink appears; then add 1 drop of

standard hydrochloric acid which should render the blank white again.

The net result of the blank determination will be a small volume of base;

a correction of an equivalent volume of acid must be added to that vol-

ume of acid used in the analysis. From the data obtained in the methyl

orange and phenolphthalein titrations calculate the percentages of sodium

carbonate and sodium hydroxide in the sample. Duplicate determinations

should check within 5 parts per 1000 from the mean.

Notes. Read the note on p. 141.

The blank consists of a precipitate of barium carbonate and an excess of

barium chloride (be sure an excess of barium chloride has been used). Therefore

Page 161: quimica inorganica cuantitativa


its reaction should be slightly acidic and a small amount of standard base must beadded to obtain a faint pink color. Since the phenolphthalein end point in the

analysis is from pink to white, 1 drop of standard hydrochloric acid is added after

the blank has been rendered pink in order to return to white. The volume of acid

corresponding to the net volume of base employed for the blank obviously mustbe added to the volume of acid used for the analysis. This calculation may be

made since the standard acid-standard base ratio has been determined already.Problem. A mixture containing sodium carbonate and sodium hydroxide and

weighing 0.2375 g. is dissolved and titrated to the methyl orange end point with

0.0960 N hydrochloric acid, 47.84 ml. being required. A second sample of

the same weight is dissolved and an excess of barium chloride is added. It is

titrated with the 0.0960 N acid using phenolphthalein as the indicator, 20.22 ml.

being necessary. A blank required a net volume of 0.06 ml. of 0.0970 N sodium

hydroxide solution. Calculate the percentage of sodium carbonate and of sodium

hydroxide in the mixture.

Solution, (a) Since the acid and the base are so nearly the same normalitythe small volume of 0.06 ml. of base for the blank is equivalent to the same volumeof acid. Thus the titration to the phenolphthalein end point becomes 20.22 + 0.06

or 20.28 ml. of standard acid.

Total alkalinity= (0.04784) (0.0960) = 0.001593 =

g. eq. wts.

NaOH alkalinity = (0.02028) (0.0960) = 0.001917 =g. eq. wts.

Na 2C03 alkalinity= 0.004593 - 0.001947 = 0.002646 =

g. eq. wts.

(0.002646) (53.00) (100)


(0.001947) (40.01) (100)


= 59.0% Na 2CO3

= 32.8%NaOlI

Solution, (b) The problem may be solved also as follows. Thinking now in

terms of "A" and "B" ml. as was done in the example on p. 125, it is obvious

that the 47.48 ml. of acid in the first titration must represent (A + B) ml. Thesecond titration neutralizes hydroxide alone so that the 20.28 ml. of acid must

represent (A B) ml. Therefore (A + B) = 47.84 ml. of standard acid and

(A - B) = 20.28 ml. of standard acid. Subtracting gives 2B = 27.56 ml. of

standard acid equivalent to the NaaC0 3 .

Knowing that 27.56 ml. of 0.0960 N acid is equivalent to the Na 2C03 and that

20.28 ml. of the acid is equivalent to the NaOH, we may proceed from this pointon in exactly the same manner as shown under (a) on p. 125, leading to the

results: 59.0% Na 2C0 3 and 32.8% NaOH.



The determination of nitrogen by the Kjeldahl method depends in its

final step upon the indirect titration of ammonia coming from the original

nitrogenous substance. Amines and amides are converted into ammoniumbisulfate by digestion with concentrated sulfuric acid; the organic sub-

stance thus is decomposed, the carbon and hydrogen being oxidized to

8 For two instructive and interesting articles see Bradstreet, Chem. Revs., 27, 331

(1940); and Oesper, J. Chem. Education, 11, 457 (1934).

Page 162: quimica inorganica cuantitativa


carbon dioxide and water while the nitrogen is converted into ammonium

bisulfate. The solution then is treated with a concentrated solution of

sodium hydroxide and the ammonia thus liberated is distilled into a

measured volume of standard acid, present in excess. Finally the excess

acid is determined by titrating with a standard alkaline solution. From the

amount of standard acid equivalent to the ammonia, the percentage of

ammonia, or of nitrogen, in the original sample may be computed.

The Kjeldahl method is applicable to the determination of nitrogen in

foods, fertilizers and the like; if the nitrogen is fixed in forms other than

amines or amides, special means must be provided to convert it into am-

monium salts. Modifications of the method are necessary if the material

contains azo, cyanide, hydrazine, nitro, nitroso and similar groups. In

such cases one may consult other works for particular methods. 9 The

oxidation of the organic compounds in the sample by hot concentrated

sulfuric acid is hastened by adding anhydrous sodium or potassium sul-

fate. This elevates the boiling point and at the higher temperature the

period of time required for the digestion is reduced. Many catalysts have

been used to accelerate the digestion, including, among others, mercuric

oxide, cupric sulfate, selenium oxychloride and copper selcnite.

Procedure. Weigh samples of between 1 and 2 g. to the nearest milli-

gram and place in Kjeldahl flasks. (Take care that none of the sample

clings to the neck of the flask. After weighing wrap the sample in a piece

of filter paper and drop into the flask, or, if this method is not employed,

be sure that any material sticking to the neck is washed down when the

sulfuric acid is poured in. The use of filter paper is objectionable in that

the digestion period is thereby prolonged due to the resistance of the paper

to oxidation.) Add 10 g. of potassium sulfate, 0.3 g. of cupric selcnite (or

2 g. of cupric sulfate) and 25 ml. of concentrated sulfuric acid.

(If nitrites or nitrates are present in the unknown they would be con-

verted into nitric acid during the reaction with sulfuric acid and lost

through volatilization. This may be prevented by a modified procedure

in which 25 ml. of sulfuric acid containing 1 g. of salicylic acid is added.

After standing a half hour with occasional shaking the nitric acid is con-

verted into nitrated organic compounds. Then 5 g. of sodium thiosulfate is

added in small portions and the solution is heated for 5 minutes. The

thiosulfate reduces the nitrated compounds to corresponding amines. The

solution then is cooled and 10 g. of anhydrous potassium sulfate and 0.3

g. of cupric selenite are added. From this point forward the regular pro-

cedure is followed. Omit the salicylic acid-thiosulfate treatment only if

it is known that nitrites and nitrates are absent.)

Swirl the mixture. Heat gently on the Kjeldahl digestion rack which

9Official and Tentative Methods of Analysis, Association of Official Agricultural

Chemists, Washington, D.C., 1945.

Page 163: quimica inorganica cuantitativa


FIG. 25. Apparatus for the steam distillation of ammonia in the Kjeldahl determi-

nation of nitrogen.

provides for removal of fumes. The digestion may be carried out on the

laboratory desk if a Kjeldahl rack is not available, provided an individual

fume eradicator is employed.10 When frothing ceases increase the heat to

the simmering point and continue the digestion until the black solution

becomes clear and light green (due to the copper of the catalyst) in color.

Then digest 10 minutes longer.

Cool the flask to room temperature and carefully add 200 ml. of water.

Shake and if necessary warm slightly until solution of the cake is com-

plete. (A gelatinous precipitate will appear if any silica is present; this

may be ignored.) The solution is now ready for the liberation and distil-

lation of the ammonia.

Make ready for the distillation as follows. With a pipet place 50 ml.

of standard hydrochloric acid which is about 0.1 TV in a receiving flask

marked to indicate a volume of 200 ml. Adjust the height of this flask so

that the end of the adapter extends just below the level of the acid.

Carry out the distillation as shown in Fig. 25, or, if equipment for a bat-

tery of distillations such as is pictured in Fig. 26 is available, this should

be used; this shows a combination digestion-distillation arrangement.

With the former apparatus steam distillation may be performed; this re-

duces the tendency toward bumping.

10 Purchase from G. F. Smith Chemical Co., Columbus, Ohio; or prepare as recom-

mended by Heston and Wood. Pack a Goccb crucible loosely with medium asbestos

moistened with concentrated sulfuric acid. Place the crucible in the mouth of the Kjel-

dahl flask using lead or tin foil as a gasket to make a tight fit. The fumes are completelyabsorbed.

Page 164: quimica inorganica cuantitativa


Fi<;. 26. Kjeldahl digestion and distillation unit. (Courtesy, E. Machlett & Son.)

Place a piece of litmus paper in the Kjeldahl flask and add 50 ml. of a

cold 50 per cent solution of sodium hydroxide. (In case steam distillation

is not employed add a gram of granulated zinc to prevent bumping.)

In adding the sodium hydroxide solution hold the mouth of the flask away

from your person so that if an unexpectedly violent reaction occurs no

injury will be suffered. Swirl the flask to mix the solutions and connect

at once the flask and the condenser with the Kjeldahl trap intervening.

If the litmus paper is not blue at this point the solution is not basic, in

which case additional base must be added.

Heat the Kjeldahl flask with steam or directly, depending on the

method of distillation and continue to distil until the volume of liquid

in the receiving flask has reached the 200 ml. level. By this time all of the

ammonia will have distilled over. (In the absence of steam distillation the

receiving flask should be lowered occasionally so that the end of the

adapter is kept only slightly underneath the level of the liquid. If too far

below the level small temperature variations could cause a suction which

might draw liquid into the condenser and even into the Kjeldahl flask.)

When the volume in the receiving flask reaches 200 ml. lower this flask

until the adapter is no longer touching the liquid. Then turn off the steam

generator (or discontinue the heating). Rinse off the adapter into the

Page 165: quimica inorganica cuantitativa


receiving vessel and titrate the contents with a standard solution of about

0.1 TV sodium hydroxide using 5 drops of methyl red as the indicator. Fromthe volume of standard base used for the titration calculate how muchacid must have reacted with the ammonia. Knowing the volume of acid

necessary for the ammonia, and the normality of the acid, calculate the

percentage of nitrogen (as such or in terms of NHs) in the sample. The

equivalent weight of nitrogen, based upon the titration of ammoniumhydroxide with an acid, is of course 14.01. Duplicate results should agreewithin 0.3 per cent.

Notes. Both the weight of the sample and the volume of standard acid used to

absorb the ammonia depend upon the nitrogen content of the sample. The in-

structor may indicate the approximate nitrogen content from which one mayestimate suitable amounts of sample and acid. Or a single preliminary run may bemade using the quantities given in the above procedure, after which, if necessary,the weight of sample and volume of acid may be varied to accommodate the needsof the particular substance to be analyzed.

The determination should be carried out in an atmosphere as free as possiblefrom ammonia fumes, since the solution used in the analysis may absorb thefumes and cause high results.

For highest accuracy a blank should be run, for example, on a 1 g. sample of

sugar, exactly as outlined for the analysis. This will correct not only for nitrog-enous substances possibly present in the reagents but also for the "normal"ammonia content in the laboratory atmosphere. Subtract the volume of standardacid neutralized by ammonia in the blank from the volume of standard acid

neutralized by ammonia in the analysis. The difference is the acid actually

equivalent to the ammonia in the sample.The time required for the digestion may vary from a few minutes to an hour

or more depending on the substance analyzed. If the contents of the flask are notclear after an hour, turn off the heat, wait a minute (do not allow the acid in the

absorbing vessel to suck back into the condenser), then quickly introduce more

potassium sulfate into the Kjeldahl flask.

A very small piece of paraffin may be added to the distilling flask to reduce


When the analysis has been completed and after the contents of the Kjeldahlflask have cooled somewhat, but while still liquid, empty the flask into the sink

half full of water. If the residue solidifies in the flask it may cause the flask tocrack. Always steam out the condenser after finishing an analysis by runningsteam from boiling water through the condenser without any running water in

the jacket. It is safest to steam out a condenser also just before beginning an


Boric acid may be used instead of standard hydrochloric acid to absorb theliberated ammonia. 11 The exact concentration of the boric acid solution need notbe known because the ammonium borate so formed can be titrated directly with astandard acid solution, using methyl red or bromcresol green as the indicator.

In this procedure the only change from that already described is the substitutingof 50 ml. of 0.6 M boric acid (practically saturated) solution for the standard

11 See Winkler, Z. angew. Chem., 26, 231 (1913); Meeker and Wagner, Ind. Eng.Chem., Anal Ed., 5, 396 (1933) ; Wagner, #>., 12, 771 (1940),

Page 166: quimica inorganica cuantitativa


hydrochloric acid used to absorb the ammonia, and the final titration with stand-

ard hydrochloric acid. The end point is reached using bromcresol green when the

solution assumes a faint yellow color.

Other Applications of Kjeldahl Method. Inorganic Nitrites or

Nitrates. Dissolve accurately weighed samples of about 0.2 g. in 100 ml.

of water. Quickly add 50 ml. of 50 per cent sodium hydroxide and 3 g. of

an alloy of aluminum, copper and zinc, called Devarda's alloy. After the

reaction, in which the nitrite or nitrate is reduced to ammonia, has con-

tinued for 5 minutes, start the distillation and carry on the procedure in

the usual manner until two-thirds of the liquid has distilled over. Titra-

tion of the excess acid in the receiving flask with standard base using

methyl red as the indicator reveals the quantity of acid equivalent to the

ammonia, from which the percentage of nitrite or nitrate may be calcu-

lated. (Absorption of the ammonia in boric acid may be employed if pre-

ferred; see the final paragraph in the above section.)

Ammonium salts, analyzed according to the Kjeldahl procedure,

simply may be dissolved in water, treated with an excess of sodium hy-

droxide and the ammonia distilled over into a measured excess of standard

acid. Obviously no digestion or reduction is necessary if no organic matter

is present and the nitrogen is initially in the form of ammonium ion.

Place an accurately weighed sample of 0.2 or 0.3 g. of the salt in the

Kjeldahl flask. Dissolve in 100 ml. of water. After connecting the receiv-

ing flask containing 50 ml. of standard hydrochloric acid, about 0.1 N,

quickly pour 50 ml. of 50 per cent sodium hydroxide into the Kjeldahl

flask, connect the flask to the condenser, swirl to mix the contents and

proceed with the distillation until two-thirds of the liquid in the flask has

distilled over. Titrate the excess acid as usual, using methyl red indicator,

and calculate the percentage of ammonia present. (Absorption of the

ammonia in boric acid may be employed if desired; see final note p. 147.)

Questions and Problems

(See also problems at end of Chapters 6 and 7)

1. Explain why the use of the smallest quantity of an indicator that will un-

mistakably reveal the end point of a titration is better than a larger amount.

2. Explain the difference in the information given by a statement of the pH of a

solution of an acid or a base, and a statement of the normality of the acid or


3. Explain the difference between the stoichiometric point and the end point of

a titration. From a practical standpoint why is it that the two seldom exactly


4. In the titrations of acids and bases of various strengths against one another

what calculation is involved for forecasting in a given case just what indicator

will detect the end point satisfactorily P What data is needed to make such


Page 167: quimica inorganica cuantitativa


5. Explain why any of several indicators may be employed to reveal the end

point in the titration of a strong acid against a strong base when both the

acid and the base are about 0.1 N. Explain why fewer indicators, if any, are

available which will disclose the end point in such a titration if the acid and

the base are used in 0.01 N or in 0.001 N concentrations.

6. A bottle of concentrated hydrochloric acid has been labeled by the manu-

facturer as containing 36 per cent hydrogen chloride by weight. Outline a

method of analysis by which the statement may be checked. Include esti-

mates of a suitable quantity of sample, the concentration of reagent used and

the precision with which measurements and calculations should be made.

7. Explain fully wjiy in the analysis of vinegar the sample should be weighed

instead of measured volumetrically.

8. Show explicitly and concisely just when, in acid-base titrations, carbon

dioxide-free water must be used, and how in certain cases the presence of

carbon dioxide will cause no difficulty.

9. When carbonate and hydroxide in a mixture of the two are analyzed by the

double-titration method, without precipitating the carbonate as barium car-

bonate, there arc two reasons, both concerned with carbon dioxide, why the

titration flask should be unstoppcred only long enough to add the successive

increments of titrating acid. Explain why this is true.

10. Whereas it is possible to have a dry mixture of sodium hydroxide and sodium

bicarbonate, these two substances in the presence of moisture, or in solution,

are said to be incompatible. Explain.

11. Suppose that an unknown solution contains cither sodium hydroxide, sodium-

bicarbonate, sodium carbonate or cither of the two compatible combinations.

Let A represent the volume of standard acid required to titrate the cold solu-

tion to the phenolphthalcin end point, and B represent the additional volume

of acid necessary to reach the methyl orange end point in a simple double

titration. Explain why the information listed in the first two columns in the

following tabfe qualitatively identifies the component or components as

listed in the third column.

12. Discuss two advantages, one of a practical nature (time saving) and one which

leads to greater precision, in the analysis for nitrogen by the Kjeldahl method,

if boric acid solution is used to absorb the ammonia instead of a standard acid


Criticize the statement: the object of any acid-base titration is to neutralize

the solution in the beaker.


Page 168: quimica inorganica cuantitativa


14. A certain solution contains 0.0987 gram equivalent weight of an acid in 987

ml. of solution. What is its normality?

15. A solution of nitric acid has a specific gravity of 1.20 and contains 32.3 per

cent HN0 3 by weight. Calculate its approximate normality.

Answer: 6.14 N.

16. Given exactly 500.0 ml. of 1.000 N acid. To what volume must it be diluted

with 0.2000 TV acid in order that the resulting solution will be 0.4500 TV?

Answer: 1600 ml.

17. A solution is prepared by dissolving together 5.000 g. each of sodium hydrox-

ide and potassium hydroxide. How many ml. of hydrochloric acid, 1.000 ml.

of which is equivalent to 0.01212 g. of benzoic acid, will be necessary to

neutralize the basic solution?

Answer: 2159 ml.

18. One hundred ml. of a solution of 0.2500 AT potassium hydroxide is placed in a

250 ml. volumetric flask and brought up to the mark with a second potassium

hydroxide solution. The resulting normality is found to be 0.2000 N. What

was the normality of the second potassium hydroxide solution?

Answer: 0.1667 N.

19. Constant-boiling hydrochloric acid which was distilled at a pressure of 755

mm. and weighing 18.0206 g. is diluted to 1000 ml. Calculate the normality of

the solution.

Answer: 0.1001 N.

20. A sample of pure benzoic acid weighing 0.4026 g. is titrated with a sodium

hydroxide solution, 34.78 ml. being required. Calculate the normality of the

alkaline solution.

Answer: 0.0948 N.

21. Fifty ml. of the sodium hydroxide solution of (20) is equivalent to how many

grams of pure potassium biphthalate?

Answer: 0.968 g.

22. An analyst has prepared a solution of hydrochloric acid and one of sodium

hydroxide. He finds that 1.000 ml. of the acid is equivalent to 0.992 ml. of

the base. A sample of pure sodium carbonate weighing 0.1600 g. is dissolved

in water and 45.11 ml. of the hydrochloric acid is added. The resulting solu-

tion is acidic. Upon titrating it with the sodium hydroxide solution 14.80

ml. is required. Calculate the normality of the hydrochloric acid and of the

sodium hydroxide.Answer: HC1, 0.1000 N; NaOH, 0.1008 N.

23. A sample weighing 1.058 g. contains sodium hydroxide, sodium carbonate and

inert material. It is dissolved and titrated with 0.4800 N hydrochloric acid.

Using phenolphthalein as the indicator, 28.50 ml. of the acid has been added

when the color changes. Methyl orange is then added and an additional 10.52

ml. of the acid is necessary to reach the second end point. Calculate the per-

centage of NaOH and Na2COa in the sample.Answer: 32.6% NaOH; 50.6% Na2COs.

24. Four samples all weighing exactly 1 g. are known to be either binary mixtures

of sodium carbonate, sodium hydroxide and sodium bicarbonate, plus inert

material, or to consist of just one of the active ingredients and inert material.

From the data below state what active component or components are pres-

Page 169: quimica inorganica cuantitativa


ent and calculate the percentage of each. The titrating acid was 0.4800 Nhydrochloric.






Ml. Acid to Phenol-

phihalein End Point

Additional ml. Acid to

Methyl Orange End Point

12.08 12.08

0.00 20.20

10.04 16.61

The fourth sample required 40.00 ml. of acid to

reach the phenolphthalein end point; then

10.50 ml. excess acid was added and the solu-

tion was hoiled. Following this the solution

was cooled in an atmosphere free from carbon

dioxide and titrated with 0.2400 N sodium

hydroxide solution, 21.00 ml. being required.

aridAnswer: (1) 61.5% Na 2C0 3 ; (2) 81.5% NaIlC0 3 ; (3) 51.1% Na,CO,

26.5% NaHC0 3 ; (4) 76.8% NaOH.

25. What must be the normality of the hydrochloric acid used to determine the

percentage purity of soda ash (crude sodium carbonate) if exactly 1 g. sam-

ples are used, in order that the buret reading will be equal to the percentageof Na 2C0 3 ?

Answer: 0.1887 N.

26. (a) A mixture consists of pure sodium carbonate and pure barium carbonate.

If 0.2500 g. of it requires 32.84 ml. of 0.1000 N acid for complete neutraliza-

tion, what percentage of each component is present in the mixture? (b) If a

0.2500 g. sample of the same mixture is titrated to the phenolphthalein end

point, how many ml. of the 0.1000 N acid will be required?Answer: (a) 34.4% Na 2C03 and 65.6% BaC0 3 ; (b) 16.23 ml.

27. The nitrogen in a sample of food weighing 2.045 g. is determined by the

Kjeldahl method. The liberated ammonia is received in 50.00 ml. of acid and

the excess acid is titrated with 18.24 ml. of 0.5062 N sodium hydroxide solu-

tion. When 50.00 ml. of the acid is titrated directly against the base 40.00 ml.

is necessary. Calculate the percentage of nitrogen in the sample.Answer: 7.38%.

28. Find the equivalent weight of an acid 1.565 g. of which neutralize 35.58 ml.

of potassium hydroxide solution, 10.00 ml. of the latter being equivalent to

0.5000 g. of pure calcium carbonate.

Answer: 44.02.

Page 170: quimica inorganica cuantitativa

Chapter 9



V>|XIDATION ii^defined as the loss of one or more electrons by an atom or

an ion; reduction is the reverse, or the gain of one or more electrons by an

atom or an ion. Neither can occur in the absence of the other; that is to

'say, if an atom or ion is oxidized there must be another atom or ion simul-

taneously reduced. The atom or ion which gains electrons and itself is

reduced causes another atom or ion to lose electrons; the former is there-

fore called the oxidizing agent or oxidant, and the latter is known as the

reducing agent or reductant.

Not all oxidation-reduction reactions are suitable for application to

quantitative analysis. All such reactions are reversible and unless the

equilibrium point lies far to one side or the other the reaction >rould not

be feasible as a basis for analysis. However, there are so many oxidation-

reduction reactions which for all practical purposes are complete that this

type of reaction finds greater application than any other in volumetric


Some oxidizing agents have a more pronounced tendency to gain

electrons than others, and some reducing agents have a stronger tendency

to lose electrons than others. Thus all oxidizing agents and all reducing

agents reveal, under a given set of conditions, a definite electron affinity.

The best oxidizing agents are those which exhibit the highest electron

affinity, and the best reducing agents are those which show the lowest

electron affinity. When a good oxidizing agent reacts with a good reducing

agent the reaction almost attains completeness, since one reagent pos-

sesses a very high electron affinity a;*d the other a very low electron affin-

ity. On the other hand when the electron affinities of the oxidant and the

reductant are not widely different the reversibility of the reaction be-

tween them will be revealed by the presence of considerable quantities of

each after equilibrium has been attained. Obviously it is necessary that

we knfcw something of a quantitative nature about electron affinities if

we a/e to deal in a sure and definite manner with reactions of this type.

Thj^ we may accomplish through consideration of electrode potentials

taken up in Chapter 10. For the moment, however, let us confine ourselves

to typical oxidation-reduction reactions which commonly are used in


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quantitative determinations, and take up the questions of the equivalent

weight, normality, stoichiometric relations in general for such reagents,

and the method of balancing the equations involved.

Equivalent Weight of Oxidizing and of Reducing Agents. It will

be remembered that the gram equivalent weight of an acid, a base or a salt

is that weight of the compound necessary to provide 1 gram atomic weight

of replaceable hydrogen or its equivalent. The gram equivalent weight of

an oxidizing agent or of a reducing agent is that weight of the compound

necessary to oxidize 1 gram atomic weight of hydrogen or to reduce 1

gram atomic weight of hydrogen ion. The oxidation of one atom of hydro-

gen or the reduction of one hydrogen ion involves a transfer of one elec-

tron. Accordingly, the equivalent weight of an oxidizing agent or of a

reducing agent is the molecular weight of the compound divided by the

number of electrons lost or gained by the molecule.

It sometimes is found that a given oxidant, for example, may undergo

reduction to different extents, depending upon the conditions of the reac-

tion. For example, the permanganate ion of potassium permanganate,

employed as an oxidant in an acidic solution, is converted to manganousion with a valence of 2+ , whereas in a neutral or basic solution it is con-

verted into manganese dioxide in which the manganese has a valence of

4+ . Obviously the number of electrons gained in the two cases is different

and thus the equivalent weight of potassium permanganate is different

depending upon the acidity or basicity of the solution. Again nitric acid,

if used as an oxidant in concentrated solution, will be reduced to nitrogen

dioxide, but used as a dilute solution it is reduced to nitric oxide or even

to ammonia. The equivalent weight of nitric acid therefore varies ac-

cording to its concentration in the solution used. Evidently then it is

imperative that one consider the particular reaction which the reagent is

undergoing, thus learning exactly how many electrons per molecule are

involved, before deciding wlmt is the equivalent weight. The knowledge

of what products will result when a certain oxidation-reduction reaction

takes place comes with actual experience or may be had from the litera-

ture, i.e., from the experience of someone else. For convenience a list of

the more common oxidizing agents employed in quantitative work is

given in Table 16, together with the product which they yield under

specified conditions.

The gain or loss of electrons by an oxidant or a reductant is ascer-

tained by noting the change in oxidation number of the atom or ion which

undergoes an electron shift. For example, in KMnO 4 the potassium is re-

garded as having a valence number of 1+ and the oxygen 2~. The product

of the number of these atoms and their valence numbers gives the total

electrical charge to be attributed to potassium and oxygen, and the

algebraic difference between these products must represent the total

Page 172: quimica inorganica cuantitativa



electrical charge for manganese since the whole molecule is electrically

neutral. This is more briefly expressed as follows:

1* 8-

K Mn 4

The manganese carries a charge of 7+ and the oxidation number of the

manganese in potassium permanganate is 7+ . In the case of potassium

dichromate we have:

The chromium here carries a total charge of 12+ and, since there are two

atoms present in the molecule, the oxidation number of chromium is 6+.

Thus, if potassium permanganate takes part in a reaction in which the

manganese is reduced to Mn++ the number of electrons gained (i.e., the

total change in oxidation number) is 5. It follows that the gram equivalent

weight of potassium permanganate is KMn04/5 = 31.61 g. In the same

manner it is evident that the gram equivalent weight of this oxidant,

when used in a basic solution where Mn0 2 is the product, is KMn04/3= 52.68 g. since the electron shift is 3 electrons per molecule of oxidant.

In the case of potassium dichromate the oxidation number of the chro-

mium is 6+ and, after reduction, the chromium is in a valence state of 3+;

thus the change in oxidation number is 3 per atom of chromium or 6 per

molecule of potassium dichromate. The gram equivalent weight therefore

is K 2Cr 2 7/6 = 49.04 g.

When sodium oxalate reacts as a reducing agent one of the products is

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carbon dioxide. The change in oxidation number for each carbon must be

1. It is evident that the total charge on the two carbons is 6+


2+ 8-

Na 2 C 2 4

and the oxidation number for carbon in oxalates is 3+ . In C0 2 the valence

of carbon is 4+; thus per atom of carbon the electron shift is 1, and per

molecule of sodium oxalate it is 2. The gram equivalent weight thus is

Na2C 204/2 = 67.01 g. Similarly sodium thiosulfate, Na 2S 2 3 , when it

acts as a reductant and itself is converted into sodium tetrathionate,

Na^Oe, must have a gram equivalent weight of Na2S 2Oa/l = 158.1 g.,

as may be seen from the following:

6~ 2+ ^0^ 12-

3 ->Na 2 84 6


The sulfur may be regarded as having an oxidation number in thiosulfate

of 2+ and in tetrathionate of 2^+ (i.e., 10+ -*- 4) ; the two sulfur atoms in

one molecule of sodium thiosulfate therefore undergo a loss of one elec-

tron and the molecular weight and the equivalent weight are the same.

Standard Solutions of Oxidants and Reductants. Direct andIndirect Analyses. The normal solution of an oxidizing or of a reducing

agent is, of course, one a liter of which contains I gram equivalent weightof the solute. As a rule standard solutions are made up in concentrations

of about 0.1 TV, though frequently solutions of somewhat higher or lower

concentrations are used. Standard solutions are sometimes prepared by

accurately weighing the oxidant or reductant, dissolving and diluting to

the proper volume in a calibrated flask. Such a procedure requires that

the substance be obtainable in a perfectly pure state. Potassium dichro-

mate, potassium bromate and potassium bi-iodate, KIOa.HIOs, are ex-

amples. More often the reagents are weighed in roughly correct amounts

and after dissolving in the approximately correct quantity of water the

solutions arc standardized by titrating against a carefully weighed stand-

ard substance. Thus a potassium permanganate solution is prepared in,

say,, roughly 0.1 N concentration, and after it has stood for several daysit is filtered and then may be standardized against a suitable quantity of

accurately weighed sodium oxalate. In all cases, of course, it is necessarythat the criteria for standards mentioned in connection with acidimetry

(p. 78) be met.

It is frequently the case that during a given standardization the ti-

trating solution reacts directly with the substance acting as the standard.

Such is true when permanganate solution reacts with oxalate, the latter

Page 174: quimica inorganica cuantitativa


being the standard. In other instances the standard employed reacts with

a second substance, added in excess, and the product of this reaction in

turn reacts with the solution to be standardized. For example, we maychoose potassium bromate as the standard for obtaining the normality of

a solution of sodium thiosulfate. A suitable amount of potassium bromate

is weighed out accurately, then dissolved and acidified, and an excess of

potassium iodide is added. This results in the liberation of a quantity of

iodine equivalent to the weight of the bromate. The iodine set free then

is titrated with the thiosulfate solution and the volume of thiosulfate

necessary is equivalent to the weight of potassium bromate used as the


As was true in calculating the normality of acids and bases, so also here,

the normality of a solution of an oxidizing or of a reducing agent is always

equal to the grams of standard equivalent to a liter of the unknown solu-

tion divided by the equivalent weight of the standard. It is useless, let it

be emphasized, in calculating the normality of a thiosulfate solution, to

compute the weight of iodine liberated from the potassium iodide by the

bromate. (See the section on oxidation-reduction calculations, pp. 159-


By no means are all oxidizing and reducing substances suitable for the

preparation of standard solutions to be used in volumetric analyses. Only

those are useful which may be standardized with high accuracy and which

maintain a constant normality for a reasonable length of time. The most

commonly used standard oxidizing solutions are those of permanganate,

dichromate, eerie and bromate ions. Thiosulfate, arsenite and oxalate

arc widely used as standard reducing solutions.

Analyses involving oxidation-reduction may be made either by direct

or by indirect methods. If a sample for analysis is brought into solution

and titrated at once with a solution of known normality we say that a

direct method has been used. Thus an iron ore may be dissolved, treated

so as to bring all of the iron into the ferrous state, and titrated at once

with a standard permanganate or eerie solution. The indirect approach

is employed if the substance being analyzed is treated with a reagent in

far greater quantity than stoichiometrically necessary. In the reaction

which results it is necessary that either (a) a substance be liberated which

may be determined by titration, or (b) the excess of the reagent added

may be determined. The latter may be illustrated by the analysis of an

ore containing manganese dioxide. An excess of oxalate is accurately

weighed and added to a solution of the ore and the amount of oxalate

"uuneeded" is determined by titrating with a standard oxidizing solu-

tion like permanganate. The difference between the total weight ofoxalate

and the weight of the oxalate which reacted with the permanganate is, of

course, equivalent to the manganese dioxide.

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In order to balance an equation of an oxidation-reduction reaction it is

necessary, first of all, to know what are the reactants and what are the

products. The former presents no difficulty since we know what sub-

stances are being brought together. If the products are unknown theymust be ascertained either by referring to some reference book or by ac-

tual experiment. The chief products of the oxidizing agents listed in

Table 16 are given in the second column of that table. The principal prod-ucts of some reducing agents have already been mentioned, viz., oxalates

yield carbon dioxide, thiosulfates yield tetrathionates. In addition, it

may be mentioned that iodides, chlorides and bromides usually are con-

verted into the corresponding halogen in analytical determinations; the

products formed with other reducing agents are given in connection with

the experiments in which they occur.

To illustrate: consider the oxidation of oxalic acid by potassium per-

manganate in the presence of sulfuric acid. The reactants are simply these

three compounds. As for the products, we know that in an acidic solution

permanganate is reduced to the manganous ion, Mn++, and that oxalates

are converted by oxidation into carbon dioxide. Thus, at once, we maywrite

KMn04 + H 2C 2 4 + H2S04= Mn++ + C02 + K+ + H+

If we wish to write a molecular (instead of a simple ionic) equation wemust conclude that the metallic substances on the right have to appear as

sulfates since (aside from the oxidant arid the reductant) that is the only

cation among the reactants. 1 Thus the molecular, skeleton equationbecomes

KMn0 4 + H 2C 2 4 + H2S04= MnS0 4 + C0 2 + K2S04 + H2

Water is one of the products, of course; this is generally true if hydrogenions are in contact with an oxidant.

After establishing the skeleton equation we may fix the coefficients of

the oxidant and the reductant, and of their products on the right, by

noting the gain and loss of electrons. We have seen already that the oxida-

tion numbers of manganese in permanganate and carbon in oxalate are

7+ and 3+ respectively, and that the valence of manganese and of carbon

as they appear after reacting (as seen on the right-hand side of the equa-

tion) are 2+ and 4+ respectively. That is to say, permanganate tends to

gain 5 electrons per atom of manganese and oxalate tends to lose 1 elec-

tron per atom of carbon. But, whereas permanganate contains only one

1Conceivably the potassium and the manganese could occur as products in the

form of their oxides. However, practically all oxides react with nearly all acids to form

the corresponding salts?so that the oxides cannot appear as products in the equation.

Page 176: quimica inorganica cuantitativa


atom of manganese, oxalate contains two atoms of carbon. Therefore

each molecule of permanganate tends to gain 5 electrons but each mole-

cule of oxalate tends to lose 2 electrons. It follows then that to secure

equal numbers of electrons for both gain and loss we must use two mole-

cules of permanganate and five molecules of oxalate.

This may be diagramed more briefly as follows:

U2 (Mn'+ + 5c -> Mn+)5 (2O+ - 2e -> 2O+


Note that these electron equations direct us (in order to have an equal gain and

loss of electrons) to use 2 atoms of manganese (thus 2 molecules of permanganate)and 5 X 2 = 10 atoms of carbon (thus 5 molecules of oxalate).

Knowing this we also see that we must obtain, on the right, 10 mole-

cules of carbon dioxide and 2 of maiiganous sulfatc. Therefore

2KMnO 4 + 5H2C 2 4 + ?H 2SO 4= 2MnSO 4 + 10C0 2 + ?K 2SO 4 + MI 2

In other words we definitely fix four coefficients when we study the

gain and loss of electrons. To complete the balancing of the equation it is

only necessary to note that since there are now 2 of potassium on the left

we must place the figure I in front of K 2SO4 on the right. The total num-

ber of sulfates on the right then becomes 3, so that the figure 3 must be the

coefficient of H2SO 4 on the left. Usually it is best to leave the coefficient

of H 2 to the very last. Here, since there are 10 hydrogens from 5H 2C 2O 4

plus 6 hydrogens from 3H 2S0 4 , the coefficient for H 2 obviously must be

8. We now have the completely balanced equation:

2KMnO 4 + 5H 2C 2 4 + 3H2S0 4= 2MnS0 4 + 10C0 2 + K2S0 4 + 8H 2O

Actually, of course, the reaction is between ions and not molecules,

so that ionic equations are generally preferred. For the reaction just con-

sidered the balancing of the ionic equation consists simply in first writing

the partial equations for the two systems, oxidant and reductant, so that

both partial equations involve that number of electrons represented bythe change in oxidation number of each:

(1) Mn04- + 8H+ + 5e -> Mn++ + 4H2

(2) C 20r - 2e - 2C0 2

Since the lowest common multiple of 5 and 2 is 10, equation (1) must be

multiplied by 2 and equation (2) by 5, yielding

2MnOr + 16H+ + lOe ~> 2Mn++ + 8H2

5C 2 4- - lOe - 10C02

Page 177: quimica inorganica cuantitativa


Canceling out the +10e and lOe and adding gives the final, balanced

ionic equation, or

2MnOr + 5C 2Or + 16H+ = 2Mn++ + 10CO2 + 8H2O


The same principles which govern calculations of acidimetry also

apply to oxidation-reduction calculations. In brief, the guiding principle

is that "equivalents are equivalent to equivalents." That is to say, when-

ever a certain amount of an oxidizing agent reacts, au equivalent amountof a reducing agent must react. It often is not even necessary to write the

equation for the essential chemical reaction (though for a full knowledgeof the complete picture this is necessary) provided one already knowsthe equivalent weights of the essential substances. The following examplesillustrate the types of calculations most often encountered.

A. Quantity of Standard Suitable for a Titration.

Problem. What weight of sodium oxalate should be weighed for standardizinga solution of potassium permanganate which is about 0.1 TV?

Solution. Assuming that a titration volume of about 30 ml. is desired, the

amount of permanganate will be (0.030) (0.1)= 0.0030 gram equivalent weight.

This then also will be the quantity of oxalate needed. The equivalent weight of

sodium oxalate is 134/2. Thus the weight in grams of sodium oxalate will be

(0.030)(0.1)(134/2) = 0.201 g.

(Compare with problem on p. 79.)

B. Preparation of Standard Solutions.

Problem /. What weight of pure potassium bromate is needed to prepare 2000

ml. of 0.05000 N solution?

Solution. Bromates, when they function as oxidants, are reduced to bromides:

the gain of electrons therefore is 6 per bromine. Since there is only one bromine in

KBrG 3 there will be a gain of 6 electrons per molecule. The equivalent weight thus

is KBrO 3/6 = 167.01/6 = 27.84. This is the weight necessary to make 1 1. of

1.000 N solution. Therefore 2000 ml. of 0.05000 N solution require

200?{ooo(0.05000)(27.84) = 2.784 g.

Problem 2. What weight of sodium thiosulfate, Na 2S 2 3.5H 20, is needed to

prepare 750 ml. of approximately 0.1 N solution?

Solution. If sodium thiosulfate as a reductant loses 1 electron per molecule

(see p. 155), its equivalent weight is2 2 = 218.2. This is the weight

necessary to make 1 1. of 1 N solution. Therefore 750 ml. of 0.1 N solution requires

approximately75?iooo(0.1)(248) = 18.6 g.

C. Calculation of Normality. (Compare calculations, p. 80, Chapter 6.)

Problem 1. A solution of potassium dichromate was standardized against iron

wire, 99.50 per cent pure. The weight of wire used was 0.2025 g. If 34.28 ml. of

the dichromate solution was needed for the titration, what is its normality?

Page 178: quimica inorganica cuantitativa



__ g- of Standard o 1 1. of unknown""

eq. wt. of standard

(0.2025) (0.995)


N = 0.1052

Problem 2. What is the normality of a eerie ammonium sulfate solution if

48.75 ml. will oxidize that weight of KHC 2 4 which requires 25.08 ml. of 0.1492 NNaOH for neutralization?

Solution, (a) This problem may be solved by application of the formula,

N.V = N' .V (see p. 82, Chapter 6). However, it must be remembered that

KHC 2O4, with its one replaceable hydrogen, reacts with NaOH as a monobasic

acid and, as an acid, its equivalent weight is equal to its molecular weight. With

eerie solutions KIIC 204 reacts as a reductant and gives up 2 electrons per mole-

cule; therefore, as a reducing agent the equivalent weight of KHC 2 4 is half

of its molecular weight. It follows, therefore, that the normality of any solution of

KHC 2O4 is twice as great when used as a reductant as when used as an acid. This

means, in turn, that the product of TV . V for KHC 2O 4 is twice the product of

N . V for NaOH, when N . V for the former is related to N . V for the eerie solu-

tion. With these points in mind we obtain

Ceric Solution Oxalate Solution

V . N = N' . V(48.75)7V = 2[(25.08)(0.1492)]

AT = 0.1535

Solution, (b) Problem C-2 may be solved in two steps as follows, first calculat-

ing the normality of the oxalate solution (assume the oxalate to be dissolved in

any convenient volume of solution, say 100 ml.), and then calculating the nor-

mality of the eerie solution.

Oxalate Solution Hydroxide Solution

Stepl. V.N = N' . V100 TV = (0.1492) (25.08)

N = 0.03742, as an acid

Therefore N of oxalate as a reductant = 0.07484.

Ceric Solution Oxalate Solution

Step 2. V . N = N' . V(48.75) AT = (0.07484) (100)

N = 0.1535

Solution, (c) Problem C-2 also may be solved from the standpoint of equivalent

weights. That is, we may compute the number of gram equivalent weights of

NaOH involved. There must be the same number of gram equivalents of oxalate

when figured as an acid, or twice that number when figured as a reductant. The

gram equivalents of the eerie oxidant must be the same as the gram equivalents of

oxalate when the latter is figured as a reductant. Knowing the gram equivalentsof eerie present per liter, we have the normality of the eerie solution.

Page 179: quimica inorganica cuantitativa


nvi r,Tj (25.08) (0.1492)G. eq. wts. of NaOH - ^ ^ -'

,T, TT^^ (25.08) (0.1492)G. eq. wts. of KHC 2 4

= - ^i ' as acid1UUU

Jirrrsirt 2(25.08) (0.1492) ,

G. eq. wts. of KHC 2 4 = j^~-> as reductant

2(25.08) (0.1492)G. eq. wts. of eerie in 48.75 ml. eerie solution = r^r > as oxidant

, . . 1AAA . ,. 1000 (2)(25.08)(0.1492)

G. eq. wts. of cenc in 1000 ml. cenc solution = -

-^ ^^p.4o. I O 1UUU

= 0.1535

Thus normality of eerie solution = 0.1535

D. Analyses.Problem 1. (a) Calculate the percentage of iron in an ore if a 0.2500 g. sample,

dissolved in acid and completely reduced, requires 30.32 ml. of 0.1000 N potas-

sium permanganate solution for oxidation, (b) What is the "iron value" of the

permanganate solution?

Solution, (a) Since 1000 ml. of 1 N solution contains 1 g. eq. wt., then

r*f\ n<

(0.1000) = 0.003032 g. eq. wt. of KMn0 4

used and therefore also equalsthe g. eq. wts. of iron present

Then(0.003032) (55.85) =

g. of Fe present and

(0.003032) (55.85) (100)

0.2500Fe ^ 67 7

Solution, (b) The term "iron value" means the number of grams of iron which

may be oxidized by 1 ml. of the oxidant. Since 1000 ml. of 1 N solution contains

1 g. eq. wt. it obviously is capable of oxidizing 1 g. eq. wt., or 55.85 g. of iron; 1 ml.

of 1 TV solution (of any oxidant used, whether permanganate or not) will oxidize

0.05585 g. of iron. But this solution is only 0.1000 N and thus its iron value is

evidently 0.005585. In general, then, the metal value is

N(eq. wt. of constituent sought)


or, in the present case,

H^l? (55.85) = iron value = 0.0055851000

Problem 2. Suppose we have the same data exactly as given in Problem D-l

except that, instead of the normality of the permanganate solution being given,

we are told that 1 ml. of the permanganate solution is equivalent to 0.006701 g.

of Na 2C 2 4 . What percentage of iron is present?Solution. The problem could be solved in two steps, first by calculating the

normality of the KMn0 4 solution, then solving as already explained. The normal-

ity of the KMn0 4 is

Page 180: quimica inorganica cuantitativa


g. of standard o 1 1. of unknown

eq. wt. of standard


However, this problem may be more directly attacked as follows:

Since the iron present is equivalent to 30.32 ml. of permanganate it is equiva-

lent to (30.32) (0.006700) g. of Na 2C 2 4 . This is equal to(30 -33) (0.006700)

lo4.U/^&= 0.003032 g. eq. wt. of oxalate, and we have the same number of g. eq. wts. of

iron present. Accordingly,

(0.003032) (55.85) =g. of Fe present

which, as already shown, equals 67.7% Fe for the 0.2500 g. sample.Problem 3. Calculate the percentage of iron in terms of Fe 2 3 in an ore if a

0.2500 g. sample, dissolved in acid and completely reduced, requires 30.32 ml.

of 0.1000 N potassium permanganate solution for oxidation.

Solution. This problem is precisely the same as D-l except that the answer is

called for in terms of iron oxide instead of metallic iron. In other words, the ques-tion is, if the iron in the ore were isolated and converted into Fe 2 3 , what percent-

age of the total weight of the ore would this Fe 2 3 constitute?

Proceeding in the same manner as under solution (a) of problem D-l, weobtain

0.003032 =g. eq. wts. of KMn0 4 used and therefore

also equals the g. eq. wts. of Fe 2 3


The equivalent weight of Fe 2 3 is the molecular weight divided by 2 since there

are two atoms of iron, each of which involves 1 electron when oxidized from the

ferrous to the ferric state. Then *


7 -. of Fe 2 3



Problem b. A certain oxidizing solution has an iron value of 0.005000. In the

determination of iron in a 0.5000 g. sample of ore 25.00 ml. of the solution was

required to oxidize the iron. What percentage of iron was in the ore?

Solution. The iron value of the oxidant is 0.005000, so that 1 ml. of the solution

will oxidize 0.005000 g. of iron. Therefore

(25.00) (0.005000) =g. of Fe present, and

(25.00) (0.005000) (100)

0.5000% Fe = 25.00

Problem 5. One ml. of a sodium thiosulfate solution is equivalent to 0.005567

g. of pure KBr0 3 . (a) Calculate its normality, (b) What is the copper* value of thethiosulfate solution? (c) What weight of copper ore must be weighed out for

analysis in order that the percentage of copper present will be identical with the

buret reading?

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Solution, (a) The equivalent weight of KBrO 3 is one-sixth of its molecular

weight (see Table 16). Therefore, for the thiosulfate solution

Solution, (b) From Table 16 it is seen that copper when reduced changesvalence from Cu++ to Cu+, or 1 electron is gained. Therefore, its equivalent weightis equal to its atomic weight so that for this thiosulfate solution

Copper value = ^^54) = 0.0127110UU

Solution, (c) In general, evidently

(Cu value) (ml. thio.)(100)% (<u = - -~---.-

wt. sample


% (ai = XThen

ml. thio. = XLet

wt. sample = YThen

Y (O.OI271)(X)(100)x-Y

Canceling out X on both sides and solving for Y,

Y = 1.271 g. of sample.

The above may appear to be a"trick" problem but it is not. In laboratories where

many identical analyses are to be made much time may be saved by such an

expedient as the above, since once the weight of sample to be taken is calculated,

further computations are avoided.

Page 182: quimica inorganica cuantitativa

Chapter 10


_1_HE STUDY of the subject of oxidation and reduction may be ap-

proached in many ways. Oxidation originally meant a gain of oxygen by a

substance; later it was said to be a gain in positive valence. Again it is

defined as a loss of electrons. One substance is called a good oxidizing

agent because it is found by experiment to be capable of effecting the

oxidation of a great many substances; another oxidant is regarded as

weak because it oxidizes relatively few substances. If we are to acquire a

quantitative and comprehensive understanding of oxidation it is neces-

sary that we learn how to measure the effectiveness of an oxidant, whatit may or may not do in reacting with reductants, and why, for example,a given substance behaves as an oxidant toward one compound but as a

reductant toward another. Such interpretations may be developed best

from the standpoint of electrode potentials.


When metals other than the alkalies and alkaline earths are broughtinto contact with water, they do not appear to dissolve. However, there

is a tendency for them to do so and the slight extent to which they suc-

ceed results in the formation of a small concentration of metallic ions in

the solution. Zinc, for example, when it dissolves, even slightly, undergoesthe reaction

Zn- 2e->Zn-H-

and the electrons set free accumulate on the undissolved metal, impartingto it a negative charge. This brings about a difference in electrical poten-tial between the solution containing the positive zinc ions and the metallic

zinc dipping into the solution.

Of course as soon as zinc ions enter the liquid they begin to deposit

upon the zinc strip, that is,

Zn++ + 2e -> Zn

and sooner or later the rate of ionization must equal that of deposition, or

Zn - 2e ^ Zn++164

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When the rates become equal a state of equilibrium is attained and for

the equilibrium at a given temperature and pressure


Since Zn is a solid its concentration is constant so that the equilibrium

expression may be simplified to read

lZn++] = K

The above mathematical expression simply states that when metallic

zinc is placed in contact with a solution of its ions there is one concentra-

tion, and one only, at which there is no difference in electrical potential

between the solution and the metal. This is called the equilibrium con-

centration, and under given conditions of temperature and pressure it is

characteristic for each metal.

Suppose that a piece of metal is immersed in a solution of its ions

not of equilibrium concentration. The net reaction which takes place will

be that which tends to establish the equilibrium concentration. Immersed

in a more dilute solution of its ions the metal will send forth more ions

into the solution and thus itself acquire a negative charge. If the metal is

placed in a solution of its ions of greater concentration than the equilib-

rium value, metallic ions will deposit from the solution, and this would

cause the metal to acquire a positive charge in respect to the solution.

In either case a reaction takes place which tends toward the equilibriumconcentration. Thus there will be a difference in electrical potential be-

tween metal and solution, and the extent of this difference will depend

chiefly on two factors, namely, just what the value of the equilibrium

concentration is, and the actual concentration of the ions in the solution.

It appears from what has been stated above that any metal could be

charged positively (by placing it in a solution of its ions of greater than

equilibrium concentration) or negatively (by placing it in a solution of its

ions of less than equilibrium concentration) or even not at all (if placedin a solution of exactly equilibrium concentration). With some metals,

like silver, the equilibrium concentration is of such magnitude that anyof these three possibilities can be realized. In the case of zinc, however, the

equilibrium concentration is so high that, due to the limit of solubility

of its salts, no solution of equilibrium concentration can be made; there-

fore, zinc always acquires a negative charge when dipped into any solu-

tion of zinc ions.

Limited Solubility of Metals. Asa matter offact, if zinc is immersedin pure water no appreciable quantity of zinc dissolves, and when silver is

placed in contact with a concentrated solution of its ions no appreciable

Page 184: quimica inorganica cuantitativa


amount of silver ions is deposited. The explanation is that either action is

brought to a standstill almost as soon as started. When zinc, for example, is

placed in water it does dissolve very slightly. The zinc strip thus acquires

a negative charge and, being negative, it attracts to it the very ions which

donated electrons and charged the strip negatively. Thus there results an

electrical double layer as sketched in Fig. 27, and the electrostatic attrac-

tion so created forestalls further solution of the zinc atoms. If additional

atoms of the metal did give up electrons the ions would be pulled back by

Zinc bar

:====!+ - -+^?-:. Zinc ions

FIG. 27. Electrical double-layer between

metal and solution.

the attraction of the negative strip of zinc and also would be repelled bythe blanket of positive zinc ions crowding around the metal. Therefore,

instead of there being two opposing forces namely, the force tending to

cause solution (called the electrical solution tension) and the force tend-

ing to cause the ions to deposit (the osmotic pressure) we actually have

the following:

Solution tension <= Osmotic pressure + Electrostatic attraction

Because the second factor on the right is so large it follows that the os-

motic pressure may be almost zero when the opposing forces are in bal-

ance. This means that almost no ions are in solution, or that practicallyno metal has dissolved when the reaction reaches a stable state. Like-

wise, a state of balance between the opposing forces comes about before

an appreciable quantity of ions are deposited upon a metal when it dipsinto a solution of its ions of greater than equilibrium concentration.

As a consequence, when a metal is in contact with a solution of its ions,

even though only the slightest quantity of metal may have dissolved (or

ions deposited), there will develop a potential between the metal and the

solution unless the concentration of ions in the solution is the characteris-

tic equilibrium concentration.

Page 185: quimica inorganica cuantitativa



The question naturally arises: What is the magnitude of the potential

which exists when a given rnetal isTn contact with a solution of its ions?

The magnitude depends upon the relative tendencies of the atoms to be-

come ions and thejjpns to redeposit as atoms. Under a given set of con-

ditions i.e., the particular metal (or any element, for that matter) and a

given temperature and pressure the electrolytic solution tension is a

constant. The tendency of the ions to redeposit is proportional to the

osmotic pressure and therefore to the concentration of the ions in the solu-

tion. Since the osmotic pressure depends on the concentration, it is neces-

sary that we adopt a standard concentration before attempting to obtain

a measure of a given electrode potential. The concentration of ions which

has been accepted as standard is that in which the activity of the ions is

unity. With this understanding the elements may be listed as in Table

17, which gives the single electrode potentials, E^,under the standard

condition of concentration and at 25. It may be shown 1 that the JS&rnst

equation2gives the relationship for the potential existing between a metal

in contact with a solution of its ions. At 25 it takes the form,3

where E is the electromotive force (the potential) in volts between the

metal and the solution, E is a constant for a given metal, n is the valence

of the ion4 and C is the concentration of the ions in the solution in moles

per liter. The Nernst equation is accurate only for dilute solutions. Actu-

ally it is the activity of the ions instead of the concentration which should

be employed, but the use of the latter does not greatly affect the value of

E in fairly dilute solutions.

Note that when C is equal to unity (standard concentration), log C

equals zero and equation (1) yields E = E. This particular value, E, is

known therefore as the molar electrode potential. (In the older literature

it has often been called the normal electrode potential.)

The single electrode potential cannot be directly measured for the

simple reason that to measure a given e.m.f. (electromotive force) it is

1 See any textbook in physical chemistry.2 Nernst, Z. physik. Chem., 4, 129 (1889).3 In all calculations in this chapter involving the Nernst equation the temperature

of 25 will be understood unless otherwise stated.4Actually n is the difference between the valence of the metal and the ion. Since

the valence of the metal itself is zero, this difference becomes the same as the valence

of the ion. However, for the potential established by an ion present in two different

valence states, the value of n becomes the difference between the two valences, e.g.,

for the system Sn+++

+, Sn++ the value of n is 2. (See equation (1), p. 176.)

Page 186: quimica inorganica cuantitativa


necessary that the current of electricity be made to flow through some

electrical measuring device such as a voltmeter or a potentiometer. This

necessitates a completion of the electrical circuit by passing the current

from one potential, through the measuring instrument, and thence to a

second metal which is in contact with the original solution, either directly

or indirectly. The second metal of course has its own potential. Thus what

may be done is to obtain the relative value of a given electrode potential

as compared with that of another selected as the standard potential. Byinternational agreement all electrode potentials are based upon the

arbitrary assignment of zero potential to the molar hydrogen electrode.

The hydrogen electrode consists of a stream of gaseous hydrogen pass-

ing over a platinum electrode terminating in the form of a platinum foil

upon which is deposited a layer of platinum black. The platinum black

adsorbs the gas and presents a blanket of hydrogen atoms to the solution

of hydrogen ions in which it is immersed. The electrode potential obtained

with this arrangement when the hydrogen gas at a pressure of 1 atmos-

phere saturates a solution of hydrogen ions at a concentration (activity)

of 1 M, is the potential arbitrarily taken as zero. (The activity of hydrogen

ions is unity in a 1.2M solution of hydrochloric acid; in a 1 M solution the

activity is slightly less than 1 but, as already mentioned, the error in-

troduced by employing a molarity of 1 instead of an Activity of 1 is so

small that we shall henceforth speak of molarity instead of activity.)

Measurement of Electrode Potential. Suppose that wo wish to

measure the electrode potential between silver and a 1 M solution of silver

ions. We may set up a cell as shown in Fig. 28. In this cell the hydrogen

electrode is called the reference electrode since the information being

sought in regard to the silver electrode is obtained through measuring

the e.m.f. which is developed by the silver half-cell in combination with

the hydrogen half-cell. As a matter of fact hydrogen electrodes are seldom

used as reference electrodes because of the great care necessary in handling

them. The most popular reference electrode is the calomel electrode. (Sec

Fig. 37 for a sketch of the calomel electrode.) However, no matter what

reference electrode is used, once its own potential is established (based

on the hydrogen electrode) any measurement made through its use is

automatically on the basis of zero potential for the hydrogen electrode.

The cell of Fig. 28 may be designated as

Ag Ag(lM)H+ (1 M)H 2 (1 atm.)


the silver electrode being positive. The reading on the potentiometer is

the net result of four different potentials: the electrode potential of the

silver electrode, the electrode potential of the hydrogen electrode and the

potentials at the two junctions of the salt bridge and the solutions. The

Page 187: quimica inorganica cuantitativa


FIG. 28. Apparatus for determining E for silver. (A) Glass tube filled with'

mercury. (G) Galvanometer. (P) Potentiometer. (S) Salt bridge (an inverted

U-tube containing a salt solution, used to provide a pathway for conducting

ions from one half-cell to the other). (X) Silver electrode. (Y) Hydrogen elec-


last two potentials, known as liquid junction potentials, are so small

(usually amounting only to 0.01 to 0.03 volt) that we shall regard them

as negligible. The symbol ||in designating a cell as above is used to indi-

cate a salt bridge the potential effects of which are disregarded.

The e.m.f. which is registered by the potentiometer is then the dif-

ference between the potentials of the two half-cells, the silver half-cell

and the hydrogen half-cell. From the Nernst equation for the former,

and for the latter,

E! = EA. + 0.0591 log [Ag+]

E 2= E H + 0.0591 log [H+]

For the cell as a whole the e.m.f. will be

If __ T^i __ TL? "p

= (EA. + 0.0591 log [Ag+])- (EV + 0.0591 log [H+])

Since the concentration of each ion is unity and E H is arbitrarily assigned

a value of zero, the above expression becomes

E = Ex - E 2= (E'A, + 0)




Page 188: quimica inorganica cuantitativa


where Ec is the e.m.f. of the cell as read from the potentiometer. In the

case of silver it is found that the reading on the potentiometer is 0.80

volt, meaning that the potential between a silver electrode and a 1 Msolution of silver ions is 0.80 volt more positive than the molar hydrogen


In a similar manner the value of E for copper may be determined;

the setup will be the same as described for silver except that the left

half-cell consists of a copper electrode immersed in a 1 M solution of cop-

per ions. The value so determined proves to be 0.34 volt. This cell is

represented:H+ (1 A/)Cu (1 M)1I 2 (I atrn.)


Similarly the value of E Zn may be obtained; it is found to be -0.76 volt.

The zinc electrode is the negative electrode. That is to say, the molar

electrode potential of zinc is 0.76 volt below that of the molar hydrogen

electrode. Incidentally, from what has been stated above it follows that

the cell,


Cu++ (1 M) ||Zn++ (1 M) |


will yield an e.m.f. equal to

Ei - E 2= 0.3J - (-0.76) = 1.10 volts

We may sum up what has been said regarding the single electrode poten-

tials of silver, copper, hydrogen, and zinc, when measured in molar solu-

tion of their ions, as follows:


Ag+ + e = AgoCu++ + 2c - CuH+ + e - ^H,Zn++ -f 2e = Zn

Oxidizing Potential (based on

molar hydrogen electrode = 0)





Other values are found in Table 17. Thinking of the reaction as being:

(Higher Valence State) + ne = (Lower Valence State), it is the custom to

give the positive sign to those electrodes which, at molar concentrations,

yield potentials that are greater (more positive) than the standard (molar)

hydrogen electrode, and to give the negative sign to those electrodes which

yield a smaller (more negative) potential than the molar hydrogen elec-

trode. This is another way of saying that the positive sign is given to anyelectrode if the electron affinity of the ion is greater than that of hydrogen

ion, as is true of silver and copper cited above. Zinc ion, however, is found

Page 189: quimica inorganica cuantitativa


Table 17


[(</), (0 and (s) in the table indicate that the electrode reaction takes place in asolution which is saturated with the gas, liquid or solid present in the system]

Electrode Reaction in Volts

F2 (g) 4- 2* - 2F-Pb++++ 4- 2c - Pb++Mn0 4

~4- 4H+ 4- 3e = MnO2 (s) 4- 2H2O

2BrO 8-

4- 12H+ 4- 10e - Br 2 (1) 4- 6H 2OMnOr 4- 8H+ 4- 5e - Mn++ -f 4II2O2C1O,- 4- 12H+ 4- 10e - C1 2 (g) + 6H2OPbO 2 (s) 4- 4H+ 4- 2e - Pb++ 4- 2H2OCQ++++ 4- e = Ce+++ (see footnote 10, p. 199)

ClOr 4- 6H+ 4- 6e = Cl~ 4- 3H 2OBrOr 4- 6H+ 4- 6e = Br~ + 3H2OCr 2O 7

"4- 14H+ 4- 6e = 2Cr++ -

4- 7H2O0, (g) + 2e - 2C1-MnO 2 (s) 4- 4H+ 4- 2e = Mn++ 4- 2H 2ONitro-orthophenanthroline-ferrous indicator

O 2 (g) 4- 411+ 4- 4e = 2H,OIOr 4- 12H+ 4- lOe - I 2 (s) 4- 6H 2OOrthophenanthroline-ferrous indicator

IOr 4- 6H+ 4- 6e = I- 4- 3H2OBr 2 (1) 4- 2e = 2Br~VO2

+4- 2H+ -f e - VO f+

4- H 2O2Hg++ 4- 2e - Hg2


2Cu++ 4- 21- 4- 2e = Cujlj (s)

Hg+^- 4- 2e - Hg (1)

Diphenylamine sulfonic acid indicator

Hg2++ + 2e - 2Hg (I)

Ag+ + e - Ag (s)

Fe-n+H- 4- c = Fe++

Diphenylbenzidine indicator

Diphenylamine indicator

C 6H 4O 2 4- 2H+ 4- 2e - H2G 6H 4O 2

H 8As04 4- 2H+ 4- 2e = HAsO2 4- 2H2OI 8-

4- 2e = 31-

I2 (s) 4- 2e = 21-

Cu+ 4- e - Cu (s)

Fe(CN) 6sa + e - Fe(CN) 6


GU+-*- 4- 2e Cu (s)

Hg2Cl2 (s) 4- 2e - 2Hg (1) 4- 2C1- (normal calomel)

AgCl (s) 4- e Ag (s) 4- Cl~

4- e - Cu+4-2c -


S (s) 4- 2H+ -

AgBr (s) 4- e - Ag (s) 4- 1











































Page 190: quimica inorganica cuantitativa


Table 17 (Continued)

to have a lessor electron affinity than hydrogen ion so that the electrode

potential for zinc is negative. Still another way of looking at these facts

is illustrated if we merely say that zinc has a greater tendency to give up

(two) electrons (and go into solution as zinc ions) that does silver or

copper.It is obvious also that since (a) oxidation is defined as a loss of elec-

trons, (b) an oxidizing agent itself is reduced when it functions as an

oxidant, (c) high electron affinity means a strong tendency to gain elec-

trons and (d) strong oxidizing agents therefore have a high electron

affinity; then strong oxidizing agents will have large positive electrode

potentials. Conversely, the strongest reducing agents will have large

negative electrode potentials. These conclusions may be confirmed by

consulting Table 17.

SIGN OF ANODE AND CATHODE. When considering voltaic cells (which generate

electrical current) and electrolytic cells (which consume electrical current), it

should be borne in mind that the terms anode and cathode designate respectively

the electrode at which oxidation and reduction take place. Thus in the copper-

zinc voltaic cell above, since zinc goes into solution according to the reaction,

Zn 2e > Zn++, zinc is the negative electrode. Because this reaction represents

oxidation of the zinc the negative electrode is the anode. By similar reasoning it

may be shown that the positive copper electrode is the cathode.

Page 191: quimica inorganica cuantitativa


On the other hand, if an electrical current is passed from an outside source

through a solution of a zinc salt at a sufficient voltage, the positive zinc ions will

migrate to and deposit upon the negative electrode. The reaction, Zn++ + 2e

> Zn, represents reduction; thus here the negative electrode is the cathode.

Whatever reaction takes place at the positive electrode proves it to be the anode.

Therefore the signs of the anode and cathode in a voltaic cell are the reverse

of what they are in an electrolytic cell. To summarize: in voltaic cells the anode is

negative and the cathode is positive; in electrolytic cells the anode is positive andthe cathode is negative.


In perhaps no other phase of chemical studies has more confusion for

the student developed, particularly in regard to keeping straight the

matter of signs and the direction of reaction, than in electrometric

measurements. It is necessary, therefore, to define carefully the conditions

under which measurements are made, and the conventions which are

employed. The following conditions are generally accepted.

1. The standard state for the concentration of ions is I M (strictly,

activity = 1). A difficultly soluble salt is said to be at standard state

when its solution is at equilibrium with undissolved solid.

2. A pure liquid or solid is automatically at standard state.

3. A gas is at standard state at a pressure of 1 atmosphere. For a gas

in solution standard state is the concentration of the dissolved gas which

is in equilibrium with the gaseous phase when the latter is at a pressureof 1 atmosphere.

4. The activity of a solid or liquid not in a pure state is the mole

fraction, i.e., moles of component divided by total moles present. Thus in

dilute aqueous solutions the mole fraction of water is practically unityand its activity is 1. In the case of a gaseous solution ("mixture") the

activity is the partial pressure in atmospheres.5. Reactions taking place in the two half-cells are written as given in

Table 17 and not the reverse.

6. In representing the cell the electrode which has the more positive

molar potential is written on the left side. (Sec e.g., the copper-zinc cell,

p. 170.) In writing the equation for the chemical reaction taking placethe substance in the left half-cell appears in its higher valence state on the

left side in the chemical equation. (For example, note the designation of

the copper-zinc cell, p. 170; the chemical reaction is written: Cu"1"* + Zn= Cu + Zn++.)

7. In obtaining the e.m.f. of a given cell the potential of the right

electrode is subtracted from that of the left electrode; i.e., E = EI E 2 .

8. A positive e.m.f. for the cell means that the electrons in the external

circuit flow from right to left and the chemical reaction will go spontane-

ously from left to right. A negative e.m.f. for the cell means that the flow

Page 192: quimica inorganica cuantitativa


of electrons in the external circuit is from left to right and the chemical

reaction goes spontaneously from right to left.

Unfortunately a number of different conventions have arisen for designating

the setup of cells in connection with oxidation-reduction potentials and potentio-

inetric titrations. For example, some writers express the Nernst equation for a

single electrode potential at 25 as E = E + log C, while others use the

form E = E '-

log C. The difference comes about in the derivation of then

equation (see any text on physical chemistry). At one stage in the derivation we

have E = log- where the two p's represent the solution tension of the

n pelectrode and the opposing osmotic pressure of the ions. If one substitutes these

0.0591 Pvalues in the equation so as to obtain E = '-

log ^ where P is the solutionn n Ctension and C is the concentration of the ions in solution, we have: E =

0.0591 . _ 0.0591 . ~ _., e

0.0591, n .

, 4 flog P log C. The value of log P is a constant for a given

n n n

metal at 25, and assigning E as this constant the equation becomes E = E

_JJog C. But if one substitutes P and C in the inverse mariner so as to


0591 Cobtain E = log ^ then the difference in potential between metal and solu-

. ,_ 0.0591, _ 0.0591, _ . . 0.0591, _

tion becomes E = log C log P. By assigning log P asn n n

equal to E the equation takes the form E = E + log C.

Furthermore, E will be either positive or negative depending upon whether

the reaction is considered as Men+ + ne > Me , or as Me > Men+ + ne.

Again some writers give the e.m.f. of a cell as Ei E 2 while others prefer

EI + E 2 , the algebraic sum of the two single electrode potentials. But either form

leads to the same result provided one is consistent in evaluating E t and E 2 .

Still again, there are those who like to sketch the diagram of a cell in such a

manner that the flow of electrons is always from right to left in the external cir-

cuit. There appears to be no intrinsic virtue in showing the flow of electrons al-

ways from right to left, or from left to right for that matter. The really important

thing is that one always be able to reckon the direction of flow, whatever it be,

so that the polarity of the cell is revealed.

Because of such variations in custom it is imperative that a set of conventions

be chosen and thereafter consistently be followed. Any set of rules, if funda-

mentally sound, will lead to the same results and conclusions as any other set,

provided only that once accepted they are consistently applied.


In the examples of single electrode potentials already mentioned a

metal, or hydrogen, in a zero valence state was placed in contact with a

solution of its ions. The reaction occurring

Page 193: quimica inorganica cuantitativa


Me"+ + ne & Me

was in each case an example of oxidation-reduction since there was an

electron transfer, and in each case there was an electrode potential de-

veloped between metal and solution.

In the case of the hydrogen half-cell a noble metal like platinum con-

stitutes the electrode, for at the surface of the platinum we have (ad-

sorbed) hydrogen atoms surrounded by a solution of hydrogen ions. Adifference in electrical potential also is set up if a noble metal dips into a

solution containing ions which may exhibit two finite valences, e.g.,

Fe+++ and Fe++ . A cell may be arranged by combining such a half-cell

with a hydrogen half-cell, that is,

PtFe+++ (1 M)Fe++ (1 Af,

H+ (1 M)H 2 (1 atm.)


The e.m.f. of the above cell is found to be 0.78 volt. Therefore, since both

half-cells are at standard state, E Fe= 0.78.

Again we might set up the cell,


(1M)(1M) H2 (1 atm.)


and we should find that the e.m.f. is 1.45 volts. Since both half-cells are

at standard concentrations this is the value of E Ce , i.e., 1.45.

If now we set up the cell,

Pt (I M) II Fe+++ (1 M)Ce+++ (1 M) I Fe++ (1 M)


we have, since standard concentrations are specified, for the half-cell


CC++-H- + e = Ce+++, E Cc = 1.45

Fe+++ + e = Fe++, E Fe = 0.78

and upon subtracting the latter from the former.

Fe++ = Ce+++ + Fe+++, Ec. - E Fe= 0.67

ff these two solutions are connected by a salt bridge internally, a current

of 0.67 volt will pass through the external circuit. The flow of electrons

will be from right to left. This conforms with the fact that since the e.m.f.

is positive the left-hand electrode is positive, that is, the cerium half-cell,

and the electrons pass to the pole of higher positive charge. Also we note

that since in this case the CQ++++, Ce+++ system has the higher potentialit is the oxidant, and the Fe+++, Fe++ system is the reductant. Or we maysimply say that, for the concentrations specified above, ferrous ion will be

oxidized by eerie ion.

Page 194: quimica inorganica cuantitativa


Electromotive Force and Concentration. If the cerium-iron cell

described above is permitted to operate, the oxidation of ferrous ions and

the simultaneous reduction of eerie ions will proceed, but not indefinitely.

As the action continues the concentrations of all of the ions are changing,

the ferric ion and cerous ion increasing while the ferrous ion and eerie ion

are decreasing. Finally equilibrium conditions are established and at this

point the e.m.f. of the cell, which has been decreasing as the cell oper-

ated, becomes zero. Obviously there must be a definite relation between

e.m.f. and concentrations. There is, and the relation is obtained from the

Nernst equation. For each of the two half-cells the Nernst equation states

in general that

E = E. +

where [H.V.S.] and [L.V.S.] are concentrations of the ion in its higher va-

lence state and its lower valence state respectively, and where n is the

number of electrons involved when the ion passes from one valence state

to the other.

The e.m.f. of the cell, of course, will be the difference between the two

single electrode potentials, or EI E 2 , which, in general, yields

where we are thinking of the chemical reaction as being:

X (high valence) + Y (low valence)> X (low valence) + Y (high valence)

If n has the same value for both ions, equation (2) becomes,

(3) EC = Ex - E 2= Ei - E 2

0.0591 [H.V.S. (ion Y)][L.V.S. (ion X)]~10g

[L.V.S. (ion Y)][H.V.S. (ioif X)J

From the general formulations of equations (2) and (3) we may conclude:

1. If standard concentrations prevail for all concentrations within

the brackets, or if the ratio of the numerator to the denominator is unity,

then the e.m.f. of the cell is equal to Ei E 2 ,since log 1 = 0.

2. None of the values within brackets may be zero, for then the

logarithmic expression would become infinite. This means that there must

always be a finite value for each of the several concentrations, though

Page 195: quimica inorganica cuantitativa


under certain conditions some of them may be very small. In other words,oxidation-reduction reactions are reversible.

To illustrate the use of equation (3) suppose the concentrations in the

cerium-iron cell are

[Ce++++]= 10- 5

[06+++] = 1

[Fe+++] = 1

[Fe++] = 10- 5

the reaction being written: Ce++++ + Fe++ + 0+++ + Fe+++. Substi-

tuting in equation (3),

E = 1.45 - 0.78 - 0.0591 log

= 0.67 - 0.0591 log 10 10

E = 0.08 volt

The e.m.f. of the cell is positive; according to the conventions we have

adopted, this designates the polarity of the left electrode. Thus the

cerium half-cell is the positive side and the chemical reaction spon-taneously proceeds from left to right; that is, the iron is oxidized. Had the

e.m.f. of the cell been negative, the FC+++ would have oxidized the Ce+++ ;

this proves to be the case when the concentrations in the order givenabove for the four ions are 10~7

, 1, 1 and 10~ 7respectively. The

student should prove this by making the calculation.

The application of equation (2) for cases where n has different values

for the two half-cells may be illustrated by the cerium-tin cell.

Pt (10-20


Ce+++ (10 M)Sn++++ (10 M)Sn++ (10-

20 AOPt

As usual, we shall think of the chemical equation as though the left half-

cell contains the ion acting as the oxidant, as is the case for standard con-

centrations. Whether it actually will be the oxidant or, instead, will be thereductant at the concentrations given, will be answered of course when wecalculate the e.m.f. of the cell. We therefore write the chemical equation:

2CC++-H- + Sn++ <= 2Ce+++

Substituting in equation (2),

E. =(1.45

+ o. M1

= 1.45 - 1.24 - 0.15 - 0.62

Ec= -0.56 volt

Thus, since the e.m.f. is negative, the left-hand electrode is negative andthe spontaneous chemical reaction taking place as the reaction proceeds

Page 196: quimica inorganica cuantitativa


toward equilibrium is from right to left. That is to say, at these concen-

trations the Sn 11 ' ' ion oxidizes the Ce+~H'

ion, whereas at standard con-

centrations the reverse would be true.

Calculation of Equilibrium Constant. From equation (1) it is

obvious that the single electrode potential for the cerium electrode is

(4) E! = Ec. + 0.0591 log lgwhich also may be written

and the single electrode potential for the tin electrode is

Now, solving for Ecell , |


E = Et - E2 = E'c. - EO*. --0591




The logarithmic expression, it will be noted, takes the form 5 of Kf for the

reaction :

Sn++ <= 2Ce+++ + 811++++

In the preceding section it was shown that with the concentrations

there employed the e.m.f. of this cell was negative; at standard concen-

trations for all four ions it is obvious that the e.m.f. of this cell will be

Ei E 2= +1.30 volts. Hence there must be concentrations some-

where in between these at which the e.m.f. would be zero. We may ob-

tain the value of the equilibrium constant, therefore, if we set E =0, and

solve for the value of the logarithmic term thus:

= E'c. - EO^ - -,U & c. & an

2 log




[Sn^^][Ce-^] 2 1.30s



8 Note that the logarithmic expression resembles the equilibrium constant. Actu-

ally it equals Kt only when the cell reaction is at equilibrium, i.e., when the actual

concentrations are the equilibrium concentrations.

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This large value means of course that the equilibrium point for the oxida-

tion of Sn++ by 06++++ lies far to the right, that is, the oxidation is

practically complete.

Approach Toward Completeness of Reaction. Having evaluated

the equilibrium constant for a given reaction, it is possible to calculate

how nearly complete the reaction is when equilibrium has been attained.

Suppose that Sn+4 is titrated with Ce++++ solution and that the tin

solution originally contained 5 millimoles of Sn++ and that the volume, at

the stoichiometric point, is 250 ml.

First it is necessary to evaluate the four ionic concentrations as


Let [Sn++] = X at the stoichiometric point.

[Sn++++] = %5o = 2 X 10~ 2millimoles/ml. = moles/1., for prac-

tically all of the tin is oxidized at the stoichiometric point.

[Ce+++] = 4 X 10~ 2, since the concentration of the Ce+++

formed must be twice that of Sn++++.

[Ce++++] = 2X, since when chemically equivalent quantities have

been brought together there are, for each unoxidized Sn++ ion,

two unreduced Ce++++ ions.

Substituting the above concentrations in the equilibrium expression,


=ne A

we obtain

(2 X 10~ 2)(4 X 10- 2

)2 _ R 1048


* X U

Solving for X,

AYS - (2 X 10-2)(4 X 10- 2


8 X 1043

X = 5 X 10~17 moles per 1. (= 1 X lO" 15mg. Sn++ in 250 ml.)

This is the amount of tin which escapes oxidation at the stoichiometric

point. Evidently, for all practical purposes, the reaction may be con-

sidered complete.


We may follow the change in potential established by the indicator

electrode immersed in the titrating solution for example, in the titra-

tion of stannous tin by eerie cerium by calculating several values of the

potential during the titration. Let it be emphasized that what we are now

computing is not the e.m.f. of the cell which functions during the titra-

tion. The e.m.f. of the complete cell is EI Ea and at a given point in

Page 198: quimica inorganica cuantitativa


the titration the e.m.f. will have different values if a calomel electrode,

the quinhydrone electrode or the hydrogen electrode, etc., completes the

system of which the cerium-tin, platinum electrode is the other part. On

the contrary, we are now about to calculate the changing potential of the

cerium-tin, platinum electrode itself. The potential established after the

addition of various amounts of oxidizing solution from the buret is a

function (up to the stoichiometric point) of the stannic-stannous ratio.

For example, in titrating 100 ml. of 0.1 TV stannous tin with a 0.1 TV solu-

tion of eerie ion the potential after the addition of 9.09 ml. of the oxidant

is calculated as follows.

17 A T R L '05911

9 -09L = 0.15 +


-9m(Note that the ratio of the milliliters is the same as the ratio of the con-

centrations since the volume is constant.)

E = 0.15 +~~ log 0.10

E = 0.15 - 0.03 = 0.12 volt ,

After the addition of 90.91 ml. of eerie solution we have

17 Air, -05911

90 -91E = 0.15 +

2 -log 09

E = 0.15 + 0.03 = 0.18 voltv

After 99 ml. of eerie solution

T- A ic .0.0591, 99

E = 0.15 H--2~~ log y

E = 0.15 + 0.06 = 0.21 volt /

After 99.9 ml. of eerie solution

0.15 + log

E = 0.15 + 0.09 = 0.24 volt

Before calculating the potential at the stoichiometric point note that

the chemical equation for the titration reaction is

Thus at the stoichiometric point

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MgJ =





[Ce*++]_ [Sn++J~~ S tnat

Now at the stoichiometric point ECe = E8n = EKtoich.pt.

From equation (1) we have

(6) Eco = Ec. + 0.0591 log^^3Similarly,

m E..EV + !* loelf^p

By doubling equation (7) and adding equation (6) we obtain 8

(8) SEstorh pt= E Co + 2EHn + 0.0591 log L^

Since from equation (5)


it follows that the logarithmic term of equation (8) is equal to unity and

the logarithm of that term is zero. Accordingly, equation (8) becomes

3E8toich. pt.

= EVc + 2E Sn

or ,,

(9) E*.* = - = 0>57 volt

For points past the stoichiometric point the potential is calculated

from the ratio of [Ce++++] to [Ce+++] combined with E Ce , 1.45 volts.

Thus after 100.1 ml. of eerie solution

6 For reactions in which only one ion of each reactant is involved the electrode

potential at the stoichiometric point is, of course, half of the sum (i.e., the mean)of the two E values. For details of the situation when ferrous iron ion is titrated witheerie ion, see Kolthoff and Sandell, Textbook of Quantitative Inorganic Analysis, TheMacmillan Co., New York, 1943, p. 487.

In general for a reaction: aOxi + bRed2 *=* aRedi + bOx2 the potential at the

stoichiometric point is:

_ bE ! + aE 2


fl + b

Page 200: quimica inorganica cuantitativa


E-Et. + 0.0591 log

E = 1,45 + 0.0591 log y= 1.27 volts

In a like manner the potential may be calculated for additions of 101 ml.,

110 ml. and 120 ml. of eerie solution with resulting values of 1.33, 1.39

and 1.41 volts respectively. These data are plotted in Fig. 29.

The potential increases sharply in the neighborhood of the stoichio-

metric point. As a result the stoichiometric point may be ascertained in

potentiometric titrations by plotting the potential (usually the e.m.f. of

FIG. 29. Change in potential in titration of 100 ml. of 0.1 W stannous tin with0.1 N eerie solution at 25.

Page 201: quimica inorganica cuantitativa


the cell; see below) against milliliters of solution added from the buret

and dropping a perpendicular from the steepest point on the curve. The

number of milliliters so obtained on the abscissa gives the end point of

the titration. Done in this manner no color indicator is needed.

Let it be repeated here that the values in volts plotted in Fig. 29 represent the

single electrode potential between the platinum electrode and the titrated solu-

tion. The cell e.m.f. read on the potentiometer will depend not only upon these

values but, in addition, upon what constitutes the other electrode. If the molar

hydrogen electrode comprises the other half-cell then the numerical values givenabove for the ceric-stannous titration would remain as they stand since the molar

hydrogen electrode has a potential of zero. If, however, the calomel electrode

were used in the ceric-stannous titration the e.m.f. would differ from those givenabove to an extent governed by the potential of the calomel electrode itself. For

example, with the normal calomel electrode the cell for this titration is represented

HgHg2Cl 2

RC1 (1 M) ++ Pt

(The calomel half-cell is placed on the left since it has a higher positive potentialthan the tin half-cell at standard concentrations.) The e.m.f. developed is calcu-

lated from

E. - E, - E, - EOC_ - EV + log

For example, after 90.91 ml. of eerie solution has been added,

A10 0.0591. 90.91Eo = 0.13- lo*wEe = 0.13 - 0.03 = 0.10 volt

In a similar manner the e.m.f. to be expected on the potentiometer may be calcu-

lated for all additions of eerie solution throughout the titration.


The Oxidizing or the Reducing Agent May Serve as Its Own In-

dicator. If the reagent used in an oxidation-reduction titration is highly

colored and if it is converted as it undergoes reaction into a colorless

compound, the end point is revealed by the appearance of color whenone excess drop of reagent has been added. Potassium permanganate is

one of the most valuable oxidizing agents in quantitative work because

of this property. Added to a reducing solution it is converted into color-

less manganous compounds until there is practically no reductant left;

then one excess drop of permanganate gives a pink tint even in a total

volume of several hundred milliliters.

Page 202: quimica inorganica cuantitativa


Solutions of iodine when dilute have a yellow color. Titrations with

iodine or those producing iodine show the stoichiometric point by the

disappearance or the appearance of the yellow tint. As a rule, however,

starch is added to such solutions shortly before the end point because

starch is a specific indicator for iodine, yielding the familiar intensely

blue color. Ceric ions also have a yellow color, and with some practice

one may detect the end point of titrations involving this oxidant with-

out the use of a specific indicator.

Internal Oxidation-Reduction Indicators. It will be remembered

that indicators for acid-base titrations are weak organic acids or bases

which exhibit the property of changing color when the pH of the solu-

tion in which they are used is varied over a small pH range. It also will

be recalled that the proper indicator for acidimctry is one which under-

goes the color change at a pH close to that which the titrated solution

will have at the stoichimetric point.

An oxidation-reduction indicator presents a similar problem except

that we are directly concerned with the potential characterizing the indi-

cator. An oxidation-reduction indicator is a substance which has differ-

ent colors in the oxidized and in the reduced states, and which may

undergo oxidation and reduction reversibly. Just as is the case with any

oxidant, there is an oxidizing potential associated with every oxidation-

reduction indicator, and the proper indicator for a given titration is one

the potential of which coincides with that established at the stoichiometric

point by the system present in the titration.

Assume, for example, that an unknown ferrous iron solution is being

titrated by a eerie solution of known normality. At the stoichiometric

point the potential of the system will be

Ece + KF* 1.45 + 0.78 .. 10 ,.

E8toich.pt. = -^-~ - =-J

= 1-12 volts

Thus for this titration, with eerie solution used as the oxidizing agent,

the indicator should have a potential of around 1.1 volts. Accordingly

orthophenanthroline-ferrous ion, commonly called "Ferroin," shown in

Table 17 to have a potential of 1.14 volts, would be suitable and is in

fact much used for this titration. It changes in color from red to pale blue

when oxidized.


number of these indicators have come into use ranging in potential from

around 0.25 volt to about 1.3 volts. 7 In addition to "Ferroin/' just men-

tioned, the following are typical of those most widely used.

7 See Kolthoff and Stenger, Volumetric Analysis, Vol. I, Interscience Publishers,

New York, 1942, p. 140; and Kolthoff and Sandell, Textbook of Quantitative Inorganic

Analysis, The Macmillan Co., New York, 1943, p. 496.

Page 203: quimica inorganica cuantitativa


Diphenylamine and diphenylbenzidine are only slightly soluble in water

but they do dissolve in sulfuric acid. Made up in solutions containing1 g. per 100 ml. of concentrated sulfuric acid they may be employed a

few drops to a titrated solution of 200 to 300 ml. At an oxidation potentialabove 0.76 volt the color is violet; below 0.76 volt, colorless.

Diphenylamine Sulfonic Acid. The sodium salt is usually used. Thecolor change occurs at a potential of 0.84 volt. In the reduced form it is

green, whereas the oxidized form is reddish violet. It is prepared as

0.2 to 0.5 per cent solutions of the salt in water. It may be used in the

titration of ferrous iron ; after the end point is reached the color changes

slowly on standing. While neither diphenylamine nor diphenylbenzidine

can be used as indicators in the presence of tungstates, since they form

insoluble tungstates, diphenyJamine sulfonic acid can be used in such a

solution. This is an important advantage since iron ores often contain

tungsten compounds. Another advantage of this indicator is that the

color is so pronounced that the change may be observed when dichromate

is the oxidant in spite of the green chromic ion color which that oxidant

produces when it is reduced.

Similar to"Ferroin," as far as color change is concerned, is the ferrous

ion complex of nilro-orthophenanthroline. This indicator is known as* 4

Nitro-Ferroin" and has a high oxidation potential of 1.25 volts.


1. (a) Calculate the e.m.f. of the cell

PtCe++++ (10-

7 M) II Fe+++ (1 M)M) Fe++ (10~

7 M)Pt

(b) Which electrode is positive? (c) In what direction will the chemical reaction

proceed ?

Answer: (a) ECeii= 0.16 volt; (b) positive electrode is Fe+++, Fe++ ; (c)

reaction goes spontaneously from right to left, i.e., Fe4"*"*" oxidizes Ce+++ .

2. (a) Calculate the e.m.f. of the cell

PtFe+++ (0.1 M)

\\Sn++++ (0.01 M)

Fe++ (0.01 M) [|Sn++ (0.1 M)


(b) Which electrode is positive? (c) In what direction will the chemical reaction

proceed ?

Answer: (a) Eceu = 0.72 volt; (b) positive electrode is Fe+++, Fe++; (c) reac-

tion proceeds spontaneously from left to right, i.e., Fe+++ oxidizes Sn++.

3. From the general equation for the single electrode potentials for the ceric-

cerous system and for the ferric-ferrous system, and from the values of Ece andE Fe taken from Table 17, calculate the equilibrium constant for the reaction:

Fe++ <= Ce+++ + Fe+++

Answer: 2.2 X 1011.

Page 204: quimica inorganica cuantitativa


4. (a) Calculate the equilibrium constant for the reaction:

MnOr + 8H+ + 5Fe++<=* 5Fe+++ + Mn++ + 4H 2

(b) A solution containing 5 millimoles of Fe++ is titrated with 0.5000 N KMn0 4

solution. If the final volume is 250 ml. what weight of iron remains unoxidized

at the end point if 0.03 ml. excess KMn04 solution is added and the hydrogenion concentration is 1 M? (Remember that the equation for the electrode

potential of the ferric-ferrous system must be multiplied by 5 in order to bal-

ance the over-all chemical equation, but that EF nevertheless remains 0.78.

Note too that the mole fraction of water is practically unity so that we maysubstitute 1 as the activity for water.)

Answer: (a) 8.5 X 1060; (b) 5.8 X 10~ 13

g. of Fe++ in 250 ml.

5. From the oxidation-reduction potentials from Table 17 for H 2 + e = 3^H 2

+ OH-, E = -0.83 and for H+ + e = ^H 2 ,E = 0.00, show that the ion

product of water at 25, Kw , has a value of approximately 10~ 14. (Remember

the activity of water = 1.)

6. What is the electrode potential at the stoichiometric point when a solution of

ferrous iron is titrated with a solution of eerie ionP

Answer: 1.12 volts.

Page 205: quimica inorganica cuantitativa

Chapter 11




JL OTASSIUM permanganate solution is probably the most widely used

oxidarit in quantitative analysis. This is largely due to three advantages.

First, potassium permanganate has a high oxidation potential (see Table

17) and therefore oxidizes a great many reductants. Second, the reactions

between permanganate ion and the majority of reducing substances pro-

ceed rapidly. Third, due to the fact that permanganate solution is deeply

colored, this oxidant acts as its own indicator, one excess drop of the

0.1 ^V solution imparting a distinct color even to 300 ml. of solution.

On the other hand, this reagent has some disadvantages. It cannot be

obtained in a perfectly pure state, even the highest grade containing

some manganese dioxide. A standard solution cannot be prepared, there-

fore, by accurately weighing out a definite quantity and dissolving in an

oxact volume of solution. Instead, a solution of approximately the desired

concentration is prepared, after which it must be standardized against a

suitable standard. Before standardizing the solution it must be allowed

to react with the small amount of organic matter present in the distilled

water. This results in the formation of manganese dioxide,

MnO 4- + 4H+ + 3e -> MnO2 + 2H 2O

and manganese dioxide catalyzes the reaction,

4MnO 4- + 2I1 2O -> 4OH- + !MnO2 + 3O 2

Obviously then, in order to obtain a stable permanganate solution for

standardization, it must be freed from organic matter and from manga-nese dioxide. Another limitation is that hydrochloric acid cannot be used

to acidify solutions which permanganate is to oxidize since the chloride

ion is oxidized to chlorine by permanganate.1 As a rule sulfuric acid is

used instead. In cases where hydrochloric acid has been employed to

bring a sample into solution, the solution must be evaporated with

sulfuric acid in order to expel the hydrochloric before titrating with

permanganate.1 Baxter and Zqnetti, Am. Chem. J,, 33, 500 (1905).


Page 206: quimica inorganica cuantitativa


In acid solution permanganate takes up five electrons when it func-

tions as an oxidant,

MnOr + 8H+ + 5e - Mn++ + 4II 2O

whereas in neutral or basic solution three electrons are involved.

r + 2H 2O + 3e -* MnO2 + 4OII-

Potassium permanganate solution is used in many quantitative de-

terminations. Among the most common are those involving the following

oxidations occurring in acid solution:

Fe++ - e = Fc+++

AS+++ - 2e =- 2e =- 2e =

V++ - 2e = V++++- e =_ =

MO+++ - 3e = MO++++++C 2O4

= - 2e = 2C0 2

H2O 2- 2e = 2 + 211+

H 2S - 2e = S + 211+

Preparation of 0.1 W Solution ofPotassium Permanganate. The

equivalent weight of potassium permanganate, used in acid solution, is

158.03/5 = 31.61. For a liter of 0.1 TV solution, therefore, weigh out

roughly 3.2 g. of the solid. Dissolve in approximately 1 1. of distilled

water and heat the solution in a large beaker until it boils. Then place

over a steam bath for 2 hours, after which let the solution stand over-

night or longer. This will allow time for oxidation of reducing agents in

the water.

Filter the solution through an asbestos mat into a glass-stoppered

bottle. The solution should be kept in the dark, or if preferred tho

storage bottle may be covered entirely with a black cloth wrapper and

equipped with a siphon. In the latter case use a two-hole stopper and

insert a calcium chloride tube containing a desiccant.

Notes. The solution is filtered to remove Mn02 which was present in the origi-

nal crystals and which was formed as a result of reduction by organic matter.

An asbestos filter must be used instead of filter paper since fibers from the

latter would introduce more organic matter into the solution.

If a siphon is attached to the storage bottle it should be equipped*with a glass

stopcock. No rubber tube connections should be used. A calcium chloride tube

plugged with cotton and then Tilled with a desiccant serves as the air inlet to the


Page 207: quimica inorganica cuantitativa


Standardization. Sodium oxalate serves as the most satisfactory

primary standard for permanganate solutions. It meets all of the criteria

for standards in general. Although the reaction between permanganateion and oxalate ion is not so simple as indicated in the following equation,

2Mn04- + 5C 2Or + 16H+ -> 2Mn++ + 10C02 + 8H2

the equation does indicate the stoichiometric relationship involved.

Many studies have been made of the reaction. 2

Procedure. Dry some pure sodium oxalate for 2 hours at 110, cool

and place in a glass-stoppered bottle which is kept in a desiccator. Weighsamples of about 0.2 g., transfer to 400 ml. beakers and dissolve in 200 ml.

of 5:95 sulfuric acid. Stir until all of the oxalate has dissolved. Calculate

roughly the volume of permanganate expected for the titration and

rapidly add about 90 per cent of this volume from the buret. Allow to

stand until the solution has lost its color. Heat the solution to 55 to 60

and complete the titration by adding the permanganate slowly, beingcareful that the color has disappeared entirely before adding the next

drop. The end point has been reached when a definite pink hue whichlasts for 30 seconds is obtained. Run a blank in order to see whether or

not the diluent effect would cause an error; that is, see if* 1 drop of the

permanganate solution will color for 30 seconds 250 ml. of 5:95 sulfuric

acid. If a zero blank is not obtained, make the appropriate correction.

From the weight of the sodium oxalate and the volume of permanganateused calculate the exact normality of the solution. The relative meandeviation should riot exceed 1 part per 1000 for duplicate results.

Notes. The reaction between permanganate and oxalate is catalyzed by the

manganous ion. Thus the permanganate color fades slowly after the initial addi-tion but after further additions the color fades rapidly because the reaction pro-duces its own catalyst. In other words, the reaction is autocatalytic.

Between 55 and 60 the reaction proceeds rapidly. Above 60 there is somedanger of decomposition of oxalic acid which has been formed from the combina-tion of hydrogen and oxalate ions: H 2C 204 -> H 2 + C0 2 + CO.


Arsenious Oxide. Arsenious oxide can be obtained as pure as 99.97

per cent As2O 3 . It serves as an excellent standard for permanganate withwhich it reacts rapidly and quantitatively when catalyzed with a trace ofiodide ion. Chloride ion does not interfere.

Procedure. Accurately weigh out portions of about 0.2 g. of purearsenious oxide a^d transfer to 250 ml. beakers. Dissolve in 10 ml. of3 N sodium hydroxide which is free from oxidizing or reducing agents

3 Fowler and Bright, J. Research Nail. Bur. Standards, 15, 493 (1935); McBride, J.Am. Chem. Soc., 34, 393 (1912) ; Launer and Yost, ibid., 56, 2571 (1934).

Page 208: quimica inorganica cuantitativa


and add 75 ml. of 1:10 hydrochloric acid. Add a drop of 0.002 M po-

tassium iodide as a catalyst. Titrate with the permanganate solution

until finally 1 drop causes a pink color which persists for 30 seconds or

longer. The end point may be secured, if preferred, by using a drop of

0.02 M orthophenanthroline-ferrous compound as an indicator. The rela-

tive mean deviation should not exceed 1 part per 1000.

Iron Wire. Pure iron may serve as a primary standard for perman-

ganate solutions, but care must be taken to obtain a pure sample and

to see that it is bright and clean, that is, free from traces of rust.

Procedure. Clean with fine emery paper a length of pure iron wire

which will weigh about 0.25 g., then wipe first with a piece of moist

Kleenex and then with a dry piece. Weigh the wire accurately. The iron

must be brought into solution and converted completely into the ferrous

state before titrating. Depending upon whether the reduction is to be

accomplished with the Jones reductor or by the Zimmermann-Reinhardt

method, the wire is dissolved either in sulfuric acid or in hydrochloric

acid. If the Jones reductor method is to be used dissolve the iron by

heating gently with 10 ml. of 1 M sulfuric acid, after which proceed with

the reduction and titration as described under the procedure on p. 191

beginning with the sentence: "Cool the solution arid dilute carefully to

100 ml. ..." (Of course no filtration will be necessary here.) If the

Zimmermann-Reinhardt method is preferred dissolve the wire by gentle

warming in 30 nil. of 6 N hydrochloric acid and proceed with the reduc-

tion and titration as described under the procedure on p. 191 beginning

with: "To the hot solution add dropwise a freshly prepared solution of

stannous chloride ..."


The most abundant iron ores are lirnonite, Fe 2 3.:rH20; hematite,

Fe2O 3 ; and magnetite, Fea0 4 . These ores contain varying quantities of

silica and small amounts of other substances including, sometimes,

organic matter. Three steps are involved in the determination of iron

with permanganate: the solution of the iron in the ore; the conversion

of all the iron to the ferrous state and the prevention of interference of

other reducing ions; and the titration of the ferrous ions with the stand-

ard permanganate solution.

The sample should be finely ground and if organic matter is presentit should be roasted at a dull red heat in a porcelain crucible for 10

minutes. Iron ores usually do not dissolve in sulfuric or nitric acids, but

as a rule are easily brought into solution with hydrochloric acid*, especially

if a little stannous chloride is added. A residue of silica remains and if it is

colored the solution should be filtered and the residue fused with sodium

carbonate. The fusion then is treated with acid to bring its iron into solu-

Page 209: quimica inorganica cuantitativa


tiori and this solution is added to the original. Ores not responding to

the sodium carbonate fusion should be treated with potassium bisulfate

(see p. 7).

Reduction of Ferric Ion. The reduction of ferric iron in the solution

prior to titration with permanganate may be accomplished by four

different methods: reduction with hydrogen sulfide, with_sulfur dioxide,

with stannous chloride or with a metal like zinc. It is always necessaryto remove any excess of reducing agent since otherwise the permanganatewould react with these reductants in addition to reacting with the

ferrous iron. The removal of the excess reducing agent is effected in

the first two cases by boiling the solution in an atmosphere of carbon

dioxide. Excess stannous chloride is destroyed by oxidation with mercuric

cKjoride., If zinc is used as the reductant no excess reducing agent is

introduced into the iron solution. The two last methods for reduction

will be taken up in detail.

ZIMMERMANN-REINHARDT REDUCTION. The ore is dissolved in a hydro-chloric acid solution. In the presence of ferric iron the chloride ion is

oxidized by permanganate unless a"preventive solution" 3


m^nganous ions is used to lower the oxidation potential of permanganateto a value below that necessary for the oxidation of chloride. The pre-

ventive solution also lowers the oxidation potential of. ferrous ion, byformation of the complex, Fe(P04)2

s, so that the permanganate still will

react quantitatively with the iron.4

Procedure. Accurately weigh samples of about 0.5 g. of the finely

divided, dry ore. If organic matter is suspected place the samples in

porcelain crucibles and heat to a dull red heat for 10 minutes; then cool

and place the crucibles in 100 ml. beakers. Next treat with acid as de-

scribed below and remove the crucibles by lifting with a stirring rod and

rinsing completely with a stream of water from a wash bottle. If no

organic matter is present the roasting may be omitted and the samplestreated with acid after weighing and directly transferring to 600 nil.


Add 30 ml. of 6 TV hydrochloric acid, cover the beaker with a watch

glass and heat to just below the boiling point for 30 minutes or longer,

until the silica residue is white or gray. (If the residue is still colored

after heating for an hour consult the instructor about fusing.) To the hot

solution add dropwise a freshly prepared solution of stannous chloride,

made by dissolving 15 g. of SnCl2.2H2O in 100 ml. of 1:2 hydrochloric

8Preparation of "preventive solution": Dissolve 70r g. of MnS<X4H2O in 500 ml.

of water. Add 125 ml. of concentrated H 2S0 4 and 125 ml. of 85 per cent H3PO4 . Dilute

to 1000 ml.4 See also "Theory of Preventive Solution,*' Fales and Kenny, Inorganic Quan-

titative Analysis, D. Appleton-Century Co., New York, 1939, p. 423.

Page 210: quimica inorganica cuantitativa


acid. The yellowish color of the ferric ion should now disappear. Add a

drop or two of stannous chloride in excess, but no more. Cool the solution

and pour in quickly 10 ml. of saturated mercuric chloride solution. Allow

to stand for 3 minutes. A white, silky precipitate should result. (Should

no precipitate appear too little stannous chloride was added, while the

formation of a black or gray precipitate indicates that too large an excess

of stannous chloride was present and the analysis must be started again

from the beginning.) Transfer quantitatively to a 600 ml. beaker or

Erlenmeyer flask, add about 350 ml. of water and 25 ml. of preventive

solution. Titrate at once with standard permanganate solution. The end

point has been reached when 1 drop imparts a faint pink color which

persists for 15 seconds or longer.

A blank must be run since a small but significant volume of perman-

ganate must be added after the stoichiometric point is reached to render

the color visible in such a large volume of solution. Add 2 drops of

stannous chloride to 30 ml. of 6 TV hydrochloric acid and proceed with

all of the steps of the determination. Subtract the small volume of per-

manganate required by the blank from that necessary for the analysis.

Calculate the percentage of iron in the ore. The relative deviation from

the mean should not exceed 3 parts per 1000.

Notes. The stannous chloride reduces the ferric ion: 2Fe"H"f" + Sn*4


-f- 6C1~

> 2Fe~H~ + SnCl 6". The excess stannous ion is then removed by adding mercuric

ion: Sn 4"*- + 2HgCli + 4Cl~-> SnClr + IfaCUj,. The mercurous chloride

remaining in the solution does no harm since it is insoluble and reacts with per-

manganate inappreciably in cold solutions.

Too large an excess of stannous chloride may cause the reduction of mercurous

chloride, first formed, to metallic mercury: Sn++ + Hg 2Cl 2 + 4C1~ > SnCl 6""

+ 2Hg|. This renders the analysis worthless since mercury slowly reduces the

permanganate during the titration.

The solution should be cooled to room temperature after reduction with stan-

nous chloride, for oxygen in the air would oxidize much of the ferrous ions if the

solution remained hot.

REDUCTION BY THE JONES REDUCTOR. If the sample can be dissolved

in sulfuric acid, so much the better; if hydrochloric acid must be used to

bring the sample into solution and if permanganate is to be the oxidant

used in the titration, then the solution is fumed with sulfuric acid to

expel HCL Following the fuming the solution is diluted and passed

through the Jones reductor which converts the ferric ions to the ferrous

state. The solution then may be titrated without the use of preventivesolution since no chloride ion is present. Only vanadium and molybdenuminterfere with iron determination by the Zimmermann-Reinhardt method,

but, in addition, titaniumvand chromium introduce error when the Jones

reductor method is used.

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THE REDTJCTOR. Zinc is the metal most commonly used for the re-

duction of ferric ions, though cadmium and aluminum are often used.

Zinc may be obtained free from iron but it should be tested for purity.

Amalgamated zinc is preferred to the metal alone since the amalgamation

prevents the liberation of hydrogen because of the high overvoltage of

hydrogen on mercury (see p. 305).

FIG. 30. The Jones reductor. (A) As-

bestos mat. (B) Perforated disk. (C) Wadof glass wool. (D) To suction. (E) Zinc


The Jones reductor 5 consists of a glass tube about 2 cm. in diameter

and some 25 to 30 cm. in length. It is enlarged or flared at one end and is

equipped with a stopcock at the lower end, as seen in Fig. 30. A perfo-rated porcelain disk, such as is used in Gooch crucibles, is placed in the

reductor above a wad of glass wool, and a mat of asbestos is seated onthe disk. The amalgamated zinc is lightly packed over the supportingdisk until it fills the reductor to within 2 or 3 cm. of the top. The reductor

is connected to a suction flask as shown in the figure. The amalgamatedzinc may be prepared as follows: Treat 300 g. of 20-30 mesh, pure granu-

6 See Stone and Hume, Ind. Eng. Chem., Anal Ed. t 11, 598 (1939).

Page 212: quimica inorganica cuantitativa


lated zinc with 300 ml. of 2 per cent mercuric chloride solution acidified

with 2 ml. of concentrated nitric acid. Stir for 5 minutes, decant and

wash several times with water. The zinc should now be covered with a

layer of mercury and possess a bright luster. When the reductor has been

filled with the amalgam, wash several times with water and then pass

through 100 to 200 ml. of 1 TV sulfuric acid, never allowing (from this

point on) the top of the amalgam to be exposed to air; that is, keep the

level of the liquid in the reductor above the amalgam. Test a 200 ml.

portion of I N sulfuric acid, after it has been through the reductor, by

adding 1 drop of 0.1 N permanganate solution. If the permanganate does

not impart a pink color the reductor must be washed further with acid

until a 200 ml. portion is colored with 1 drop of permanganate. The

reductor then is said to have a zero blank value.

SILVER AS A REDUCTOR. 6 The silver reductor is similar to the Jones reductor ex-

cept that metallic silver acts as the rcductant instead of amalgamated zinc. This

reductor has been employed for the determination of iron by eerie sulfate. The

silver reductor has certain advantages over the Jones reductor. In the first plaee it

shows a zero blank day after day for months. Furthermore, it does not reduce

titanium which, therefore, is not an interfering inetal in the analysis of ores if they

are reduced with silver. In addition, low concentrations of nitric acid and of am-

monium ion present no complications.

The silver reductor may be set up in a manner similar to the preparation of

the Jones reductor. A reductor tube 2 em. in diameter and 10 to 15 cm. in length

suffices. To prepare the silver dissolve 30 g. of silver nitrate in 500 ml. of water

and acidify with 5 drops of nitric acid. Then immerse a strip of pure sheet copper

having about 100 sq. cm. surface area. The silver will be thrown out of solution.

Stir occasionally. Finally when practically all of the silver has been precipitated,

wash the spongy mass with 1:20 sulfuric acid, place the metal in the tube and

wash again with 200 ml. of N hydrochloric acid. Though exposure to air causes

no finite blank, the column of silver should be kept covered with liquid. Alter

repeated use the reductor accumulates silver chloride but it is quickly regener-

ated by filling with 1:20 sulfuric acid and dipping a rod of zinc into the mass of

silver. The silver chloride is soon reconverted to metallic silver.

Procedure. Accurately weigh samples of iron ore of about 0.5 g. and

treat exactly as described on p. 191, up to the point where stamums

chloride was added. Do not add stannous chloride. To the resulting ferric

chloride solution add carefully 5 rnl. of 18 M sulfuric acid and evaporate,

without boiling, until all the hydrochloric acid has been volatilized ; that

is, until sulfuric acid fumes are noted. Cool the solution and dilute care-

fully to 100 ml. with water, then filter and wash the paper and residue

five times with 1:20 sulfuric acid. Run the solution through the Jones

reductor under gentle suction at a rate of about 30 ml. per minute (a fast

flow of drops but not streaming). Without exposing the amalgam to air

Walden, Ilammett and Edmonds, J. Am. Chem, tfoc., 56, 350 (1934); Linganeand Meites, ibid., 69, 277 (1947).

Page 213: quimica inorganica cuantitativa


follow the ferric ion solution with a wash of 200 ml. of 1 N sulfuric acid,

and then with 100 ml. of water. Remove the suction flask containing the

iron solution and titrate immediately with standard permanganate solu-

tion. As usual, the end point has been reached when 1 drop of perman-

ganate imparts a faint pink color which persists for at least 15 seconds.

Calculate the percentage of iron in the ore. Duplicate determinations

should show a relative deviation from the mean of 2 parts per 1000 or less.

Notes. Air must be excluded from the reductor since oxygen will react with

hydrogen, which is always present in the reductor, to yield hydrogen perox-ide which, in turn, reacts during the titration with permanganate: 2Mn0 4


+ 5H 2 2 + 6H+ - 2Mn++ + 50 2 + 8H 20. This would cause high results for

the iron analysis.

Since acid would eventually dissolve the zinc in the reductor, the amalgamshould not remain indefinitely in contact with acid. When the analysis has been

completed the amalgam should be washed and transferred to a jar and covered

with water for storage.


Pyrolusite, a manganese ore, is crude manganese dioxide. In this

determination the indirect approach is made for the analysis. That is,

a weighed amount of the sample is treated with an excess of a soluble

reducing agent like sodium oxalate. The manganese dioxide oxidizes an

equivalent quantity of the oxalate. The remaining oxalate is then titrated

with standard permanganate solution. Knowing the total amount of

oxalate added and the amount reacting with the permanganate, we maycalculate the quantity of oxalate needed for reaction with the sample.With these data the percentage of MnO 2 in the ore may be calculated.

This method of analysis also may be applied to the analysis of the

higher oxides of certain other metals for example, to Pb0 2 , to Pb 2 3

(PbO 2.PbO), and to Pb 3 4 (Pb0 2.2PbO). Regarding the latter two oxides

of lead, it should be noted that in each case the equivalent weight is

half the molecular weight since there is only one atom of lead in the

plumbic valence state. Upon reacting with oxalate the plumbic lead is

reduced to divalent lead; thus the whole molecule involves a gain of

two electrons.

Procedure. Weigh out accurately samples of about 0.5 g. of the dry,

finely divided ore and transfer to 400 ml. beakers. Calculate the weight

of sodium oxalate equivalent to the sample on the assumption that the

ore is pure manganese dioxide, the reaction being

Mn02 + C 2 4- + 4H+ -> Mn++ + 2C0 2 + 2H 2

Weigh out accurately about 0.2 g. in excess of the calculated amount,

using the same pure dry sodium oxalate as employed for standardizations,

and add to the sample. Add 100 ml. of 3 AT sulfuric acid, cover with a

Page 214: quimica inorganica cuantitativa


watch glass and heat gently until the evolution of carbon dioxide ceases

and no black particles may be seen. Do not permit the solution to evapo-

rate too much ; it may be necessary to add cautiously small amounts of

water a few times to keep the volume fairly constant. Finally dilute the

solution to about 250 ml., heat to 60 and titrate while hot with standard

permanganate solution. Calculate the percentage of Mri0 2 in the sample.

Duplicate results should show a relative deviation from the mean of

4 parts per 1000 or less.

Notes. During the digestion the solution should not be allowed to evaporate

to a small volume since the concentration of the solution would be increased and

the temperature might climb too high with a resulting decomposition of some of

the oxalic acid. Furthermore, concentrated sulfuric acid itself oxidizes oxalates.

Problem. A sample of pyrolusite (impure Mn0 2) weighing 0.5000 g. was

treated with 1.0000 g. of pure Na 2C 2O4 and the excess was titrated with 0.1025 TV

potassium permanganate solution, 46.72 ml. being required. Calculate the per-

centage purity of the pyrolusite in terms of Mn0 2 .

Solution. It is obvious that part of the oxalate was oxidized by the Mn0 2 and

part by the KMnO 4 . The g. eq. wts. of Mn0 2 plus the g. cq. wts. ofKMn0 4 must

equal the g. eq. wts. of oxalate. The number of g. eq. wts. of KMnO4 is'

(0.1025). The number of g. eq. wts. of MnO 2 isTrrry

where X is the weight of

Mn0 2 in grams and 43.47 is the equivalent weight of Mn0 2 (half of the molecular

weight). The number of g. eq. wts. of sodium oxalate is'

, since the equiva-

lent weight of sodium oxalate is half of its molecular weight. Therefore

1.0000 46.72((U025)+ .X

67.01 1000v ' '


X = 0.4405 g. MnO>and

1 OOX% Mn - 8{U


The method described here may be applied to any substance of which

calcium is a constituent, provided the sample is brought into solution and

freed from all other cations which would precipitate insoluble oxalates.

The procedure provides for the presence of magnesium and the alkali

metals, but no other cations. If other cations are suspected in the sample

they must first be removed. For example, if iron is present it may be

removed, along with aluminum, titanium, manganese and phosphorus,

by precipitation with ammonium hydroxide. Once the calcium oxalate

7 See article by Ungane, Ind. Eng. Chem., Anal Ed. 17, 39 (1945), for the volu-

metric determination of calcium in the presence of SiO 2 , Fe, Al, Mg, P and Mn. Themethod is accurate and rapid.

Page 215: quimica inorganica cuantitativa


has been precipitated from the sample it is treated with sulfuric acid,

converting it into calcium sulfate and oxalic acid; the latter then maybe titrated with standard permanganate solution. Thus calcium is in-

directly determined in the sense that it is actually the oxalate which is

oxidized by the permanganate; however, for each oxalate ion there was a

calcium atom present in the original sample and the results of the titra-

tion readily may be stated in terms of calcium.

Procedure. Accurately weigh out a sample containing somewhat less

than 0.1 g. of calcium (consult instructor), transfer to a 400 ml. beakerand dissolve in the smallest possible quantity of 1:1 hydrochloric acid.

If effervescence occurs during solution cover with a watch glass until

completely dissolved, then return any spray to the beaker by rinsing the

watch glass with water. Dilute to 200 ml. (If the sample consists of the

filtrate from a sample from which iron has been precipitated, beginthe analysis with the filtrate containing the calcium and follow the pro-cedure from this point on.) Add 2 or 3 drops of methyl red indicator

and 5 ml. of concentrated hydrochloric acid. Heat to boiling and add a

clear solution of 3 g. of ammonium oxalate dissolved in 50 ml. of warmwater. With the temperature of the solution at about 75 add 1 : 1 ammo-nium hydroxide dropwise while vigorously stirring the solution. When the

color changes from red to yellow stop the addition of the hydroxide. Letthe beaker stand without further heating until the precipitate of calcium

oxalate settles; then test for complete precipitation by adding a few

drops of ammonium oxalate to the supernatant liquid. After 1 hour,and no longer if magnesium is present, filter through a paper. Wash the

precipitate several times with 0.01 M ammonium oxalate; then dissolve

the precipitate with 50 ml. of hot 1 : 4 hydrochloric acid and wash the

paper with hot 1:100 hydrochloric acid. Dilute the solution to 200 ml.,

add 1 g. of ammonium oxalate dissolved in 10 ml. of hot water and heat

the solution nearly to boiling. Now reprecipitate the calcium oxalate byadding ammonium hydroxide exactly as before and allow to stand anhour or even overnight. Filter through a Gooch crucible and wash the

precipitate with cold water until the washings are chloride-free. Do not

wash more than necessary. Place the crucible and contents in a 400 ml.

beaker and add a solution of 10 ml. of concentrated sulfuric acid in

200 ml. of water. Loosen the contents of the crucible with a stirring rod

and agitate until the asbestos and the precipitate are thoroughly mixedwith the acid. Heat to 60 and titrate with standard permanganate solu-

tion. Calculate the percentage of CaO in the sample. Duplicate results

should show a relative deviation from the mean of not more than 3 parts

per 1000.

Notes. The precipitate of calcium oxalate should not be idigested too long if

magnesium is present since postprecipitation of magnesium oxalate may take

Page 216: quimica inorganica cuantitativa


place. After the reprecipitation of the calcium oxalate the digestion may be ex-

tended overnight if desired, since then the concentration of magnesium ion, even

if present originally in considerable amount, will be too low to cause appreciable


Though the above procedure serves well for the analysis of many substances

for calcium content, it is not well adapted to the complete analysis of limestone.

In order to determine silica and mixed oxides in limestone with good accuracy, a

sample of 0.5 to 1 g. is required, and this would lead to an amount of calcium

oxalate which would require around 100 ml. or more of 0.1 N permanganate in

the titration. It would be possible to dissolve the calcium oxalate precipitated

from a large sample of limestone and take an aliquot portion for the permanganate

titration, but this would be troublesome. For the complete analysis of limestone

it is customary to determine calcium gravimetrically by precipitating it and

weighing as the carbonate, oxide or sulfate (see p. 267).

A rapid method of determining calcium consists of dissolving the sample and

precipitating the calcium oxalate without prior removal of mixed oxides or silica.

The pH of the solution is fixed at 3.7 with a formate buffer. Under these conditions

calcium oxalate may be precipitated without interference from ferric or aluminum

ions, and although silica is coprecipitated this does not affect the permanganatetitration. 8

Problem. A sample of 0.2008 g. of limestone was dissolved in acid, interfering

ions were removed and calcium oxalate was precipitated. The precipitate was

treated with sulfuric acid and titrated with 0.0990 TV potassium permanganate,36.22 ml. being required. Calculate the percentage of calcium present in terms of

CaO.Solution. Evidently iCaO =e= lCaC 2O 4 o 1C 2 4

"=0= 2e. Therefore the equiva-

lent weight of CaO is half of its molecular weight, or 28.04. The g. eq. wts. of

36 22KMnOi used is --f- (0.0990) and the same number of g. eq. wts. of CaO must


be present. Therefore

36.22(0.0990) (28.04)= 0.1005 g. CaO present

% CaO = 50.1


(100) (0.1005)



In many reactions hydrogen peroxide functions as an oxidizing agent,

but in acid solution it quantitatively reduces potassium permanganatethus:

5H 2 2 + 2MnOr + 6H+ - 502 + 2Mn++ + 8H2

When the peroxide is titrated with permanganate the same initial slow

fading of the color is noticed that is experienced in titrating ferrous

solutions. As in the latter case the reaction proceeds rapidly from the

beginning if catalyzed by adding a manganous solution.

8 For details of this rapid method see Rienian, Neuss and Naiinan, Quantitative

Analysis, McGraw-Hill Book Co., New York, 1942, p. 394.

Page 217: quimica inorganica cuantitativa


Procedure. Accurately weigh about 3 g. of commercial hydrogen

peroxide and place in a 200 or 250 ml. volumetric flask. Dilute to the

calibration mark with water and then shake. With a 25 ml. pipet measure

out duplicate samples into a beaker or flask, dilute to about 100 rnl.,

add 2 ml. of concentrated sulfuric acid dissolved in 10 ml. of water and

titrate with standard potassium permanganate solution. From the rela-

tionship shown in the above equation calculate the percentage of H 2O 2

in the commercial sample. The absolute error may run as high as 0.05

per cent.

Notes. Commercial hydrogen peroxide solutions often contain small quanti-ties of stabilizers such as acetanilide which also react with permanganate. In

such a case it is preferable to determine Il20 2 iodometrieally as outlined on p.

217. In the iodometric determination the peroxide acts as an oxidant.

Peroxides of the alkali and alkaline earth metals, percarbonates and perborates

may be titrated in the same way as hydrogen peroxide if they are dissolved in

dilute acid.


The use of eerie ion as an oxidizing agent in volumetric analysis

developed rapidly after the investigations of Willard and Young, of

Furman, of Smith and others. 9 Some of the disadvantages of potassium

permanganate as an oxidant have been mentioned already. Ceric sulfate

has about the same oxidizing potential10 as potassium permanganate;

some eerie salts are even stronger oxidants than permanganate under

similar conditions. In acid solution eerie sulfate is so stable that the

normality of the solution does not change over a period of a year or more

and the solution suffers no change when exposed to light. The solution is

less subject than permanganate to decomposition when boiled. Ceric

sulfate oxidizes reducing agents quantitatively in the presence of high

concentrations of chloride ion. No intermediate products are formed

during the reduction: Ce++++ + e Ce+++. Like permanganate, eerie

sulfate can serve as its own indicator since it is yellow in color, while

the cerous ions are colorless. However, the color change is not so pro-

Willard and Young, J. Am. Chem. Soc., 50, 1322, 1334, 1372 (1928); ibid., 51,

149 (1929); ibid., 55, 3260 (1933); Furman and Wallace, ibid., 51, 1449 (1929); ibid.,

52, 2347 (1930); Furman, ibid., 50, 755 (1928); Kunz, ibid., 53, 98 (1931); Smith and

Gctz, Ind. Eng. Chem., Anal. Ed., 10, 191, 304 (1938); ibid., 12, 339 (1940); Part II of

Newer Methods of Volumetric Analysis, by Oesper, D. Van Nostrand Co., New York,1938.

10 The oxidation potential of eerie cerium assumes different values in the presenceof different anions. This may be accounted for by assuming that instead of the Ce++++

ion we are dealing with complex ions made up of cerium and the anion which happensto be present, e.g., Ce(C104)6~, Ce(N0 3)e", Ce(S0 4) 3

" and Cede". The potential as-

sociated with each complex ion would depend upon the stability of the complex.Smith and Getz, foe. cit., list the following oxidation potentials for the complex eerie

ions above in 1 to 8 normal solutions of the corresponding acids: 1.70 to 1.87; 1.61 to

1.56; 1.44 to 1.42; and 1.28 volts respectively.

Page 218: quimica inorganica cuantitativa


nounced as that of permanganate and for the majority of eerie titrations

the indicator "Ferroin" (see p. 184) is used. The color change, when

oxidized, is from red to light blue. Ceric solutions for standardization

are prepared more readily than are permanganate solutions, arid it is

obvious that readings of a buret filled with eerie solution may be made

more easily than of one containing the darkly colored permanganate.

The use of standard eerie solutions are particularly useful in potentio-

metric titrations where color indicators are unnecessary and where sta-

bility and high oxidizing power are important.

Preparation of 0,1 N Solution of Ceric Sulfate. The eerie salt most

commonly used for preparing a standard solution is eerie ammonium

sulfate dihydrate, Ce(S04)2.2(NH4)2S04.2HaO. Since only one electron

is involved when the single eerie ion is reduced the molecular weight of

this compound and its equivalent weight are identical, 632.6.

Procedure. Weigh out on a trip balance 64 g. of eerie ammonium sul-

fate and add the salt to a solution of 28 ml. of concentrated sulfuric acid

dissolved in 500 ml. of water. After the salt is in solution dilute to 1 1.

This solution is also approximately 1 TV in respect to sulfuric acid. (Prove.)

Standardization. Ceric sulfate solutions may be standardized with

arsenious oxide, sodium oxalate or electrolytic iron wire.

If the first of these is used the oxide is brought into solution with

sodium hydroxide, the solution acidified, a catalyst and the indicator are

added and the titration is carried out at room temperature. The catalytic

solution is prepared by dissolving 0.25 g. of osmium tetroxide in 100 ml.

of 0.1 N sulfuric acid. To prepare the indicator solution, Ferroin, dissolve

1.5 g. of orthophenanthroline monohydrate11 (C^Ha^.HgO) in 100 ml.

of freshly prepared 0.025 M ferrous sulfate solution.

Procedure. Weigh accurately 0.15 to 0.20 g. samples of pure, dry

arsenious oxide and transfer to 250 ml. beakers. Add 25 ml. of I TV sodium

hydroxide and warm gently to hasten solution. Cool, dilute to 100 ml.

and add 25 ml. ot6N sulfuric acid. Add 2 drops each of osmium tetroxide

and Ferroin indicator solution. Titrate with the eerie ammonium sulfate

solution until the pink color changes to pale blue. Calculate the normality

of the solution. A relative deviation from the mean of 1 part per 1000

should be attained.

Problem. Tf exactly 0.2000 g. of pure As 2 3 is used in standardizing a eerie

solution and the titration requires 40.00 ml, what is the normality?Solution. Applying equation (1) of Chapter 6 we have:

N = = 0.1010

11 This indicator may be purchased from G. F. Smith Chemical Co., Columbus,Ohio*

Page 219: quimica inorganica cuantitativa


AGAINST IRON WIRE. Weigh accurately 0.18 to 0.23 g. samples of

pure iron wire and place in Erlenmeyer flasks or in 400 ml. beakers.

Add to each sample 10 ml. of concentrated hydrochloric acid, cover with

a watch glass and heat gently'until the iron has dissolved. Rinse off the

watch glass. While still hot add stannous chloride solution until colorless,

avoiding a large excess. Cool quickly and add 10 ml. of saturated mercuric

chloride solution. A white, silky precipitate should appear; if a black or

gray precipitate results metallic mercury has been formed and the experi-

ment must be started over from the beginning. Dilute to 250 ml. with

water and add 2 drops of Ferroin indicator. Titrate with the eerie ammo-nium sulfate solution until the color changes from pink to greenish or

light yellow.

Notes. See p. 191 for preparation of the stannous chloride solution and p. 192for a discussion of the reduction of the ferric ion.

The color change of the indicator is from pink to greenish or light yellow, in-

stead of light blue, probably due to the presence of ferric ions.


Read the principle and method of the iron determination as givenunder the permanganate method for iron, pp. 190-195. The ferric solu-

tion may be reduced by use of the Jones reductor or by the Zimmermann-Reinhardt method. In the latter, after allowing the solution to stand for

3 minutes (sec p. 192), add 350 ml. of water, 20 ml. of 6 N hydrochloric

acid, 10 ml. of 85 per cent phosphoric acid and a drop or two of Ferroin

indicator. Then titrate the ferrous solution with the standard eerie am-monium sulfate until the indicator changes from pink to pale blue. Com-

pute the percentage of iron in the ore. The precision should be as high as

that obtained with the permanganate titration.

Notes. Since the eerie sulfate solution does not oxidize the chloride ion no

"preventive solution" (see p. 191) is necessary. However, the phosphoric acid

greatly facilitates the titration by forming the complex ferric phosphate ion, thus

reducing the concentration of the ferric ion. In this way the oxidation potential of

the ferric-ferrous system is sufficiently lowered to prevent the ferric ion from oxi-

dizing the indicator before the stoichiometric point is reached.

If the reduction of the ferric solution is carried out with the Jones reductor,

proceed as outlined under the permanganate method up to the point of the titra-

tion; then proceed with the titration as directed in the above section for the

Zimmermann-Reinhardt method.


Potassium dichromate has long been used as an oxidizing agent in

volumetric analysis, particularly in analyzing iron ores. It is not so

powerful an oxidant as permanganate, having an oxidizing potential of

about 1.3 volts; this is an advantage in one sense, for it will not oxidize

Page 220: quimica inorganica cuantitativa


chloride ion under ordinary conditions. In acid solution it reacts accord-

ing to the equation

Cr2 7- + 14H+ + 6e -> 2O+++ + 7H 2O

Its solution is stable and is easily prepared and it may be obtained in

such pure form that after drying at 140 it may be weighed, dissolved

and diluted to proper volume for direct use as a standard solution. 1'2

However, it is preferable to standardize dichromate solutions which are

to be used in the analysis of iron ores or alloys against pure iron wire.

With the ferrous ion it reacts as follows:

6Fe++ + Cr 2 7- + 14H+ -* 6Fe+++ + 2Cr+++ + 7H 2

Until a few years ago the titration of ferrous ions with dichromate

solution was hampered somewhat because the latter is not sufficiently

colored to serve as its own indicator, and an external indicator had to

be used. Spot tests were made from time to time as the stoichiometric

point was approached by removing a drop of the titrated solution and

mixing with potassium ferricyanide on a porcelain spot-plate. When the

spot test no longer showed the presence of the blue precipitate of ferrous

ferricyanide the oxidation of the iron was complete and the end point

had been reached. However, the advent of internal oxidation indicators

provided a satisfactory means of noting the end point of dichromate-iron

titrations without resort to the external indicator method. Ferroin indi-

cator has a rather high oxidation potential for use with dichromate, but

diphenylamine and diphenylamine sulfonic acid are very satisfactory

(see Table 17).

Preparation of 0*1 N Solution of Potassium Dichromate* Since

there are two chromium atoms in one molecule of dichromate and each

gains three electrons when acting as an oxidant, it is obvious that the

equivalent weight of potassium dichromate is a sixth of its molecular

weight. It follows that a liter of O.t TV solution will require about 4.9 g.

of the compound.

Weigh on a rough balance 4.9 g. of pure potassium dichromate, dis-

solve in water and dilute to 1 1. Place in a glass-stoppered bottle.

Standardization Against Iron. Accurately weigh 0.18 to 0.23 g.

samples of pure iron wire and place in 250 ml. beakers or flasks. Dissolve

in 10 ml. of concentrated hydrochloric acid, being careful that no loss

occurs through spray. If a beaker is used, cover with a watch glass andrinse off when the iron is in solution. Add water until the volume is

25 to 30 ml. Heat almost to boiling and add stannous chloride solution

drop by drop until the solution is colorless. Next add 1 drop excess, then

12 Willard and Young, Ind. Eng. Chem., Anal. Ed., 7, 57 (1935).

Page 221: quimica inorganica cuantitativa


cool and pour in rapidly 10 ml. of saturated mercuric chloride solution.

A white, silky precipitate should appear; a gray or black precipitateindicates metallic mercury, and in this case the solution should be dis-

carded since mercury will reduce dichromate. Allow to stand 3 minutes.

Add a solution of 250 ml. of water containing 5 rnl of concentrated

sulfuric acid and 5 ml. of 85 per cent phosphoric acid. Add 5 drops of

sodium diphenylamine sulfonate indicator (see p. 185) and titrate with

the dichromate solution. The end point has been reached when a purplecolor first appears. Calculate the normality of the potassium dichromate

solution. The relative deviation from the mean should not exceed 2 parts

per 1000.

Notes. See notes on p. 192 regarding the use of stannous chloride and mer-curic chloride.

Unless phosphoric qcid is present the color change of the indicator occurs

slowly and too soon because it is partially oxidized before the stoichiometric pointis reached. The phosphoric acid prevents this by lowering the oxidation potentialof the ferric-ferrous system through forming a complex with the ferric ions.

If the use of an external indicator is preferred, prepare a solution of potassium

ferricyanide by dissolving a small crystal in 10 or 15 ml. of water. Place a few

drops in a cup of the spot test porcelain plate and as the end point of the titration

is approached transfer 1 drop of the titrated solution to the cup with a stirringrod. The appearance of Turnbull's blue (a blue precipitate) in the spot test indi-

cates the presence of ferrous ion. Continue the titration dropwise and the spottests until no blue precipitate is obtained within 30 seconds after adding thesolution to the ferricyanide indicator.


The analysis of an iron ore by the dichromate method is carried out

in every detail as described for the permanganate process except that

no manganous solution is needed and an indicator is necessary. Follow

the procedure beginning on p. 191 through the addition of stannous

chloride and mercuric chloride. Then transfer to a 600 nil. beaker or to

an Erlenrneyer flask, add a solution of 250 ml. of water containing 5 ml.

of concentrated sulfuric acid and 5 ml. of 80 per cent phosphoric acid

and titrate as directed above in the standardization procedure. Use di-

phenylamine sulfonic acid indicator or the ferricyanide external indicator.

Calculate the percentage of iron in the sample. Duplicate results should

check within a relative deviation from the mean of 3 parts per 1000 or



(See p. 217)

Page 222: quimica inorganica cuantitativa




JLHE STANDARD oxidation potential for the iodine-iodide system as

represented by the reaction

I 2 + 2e -> 21-

stands roughly midway in the list in Table 17, p. 171. Therefore manysystems are found with oxidation potentials lower than that of iodine-

iodide, and they may be oxidized by iodine. Such reactions, when theyoccur quantitatively, may be applied to volumetric analysis by titrating

with a standard iodine solution. They are classified under the heading of

iodimetry. Among iodimetric determinations are those based upon the

following reactions:

I 2 + H2SO 3 + H2- SOr + 21- + 4H+

I 2 + H2S - S + 21- + 2H+I 2 + H 2AsO3

- + H 2O -* H2AsO4- + 2I~ + 2H+

I, + H 2SbO3- + H 2O - H2SbO4

- + 2I~ + 2H+II C? 4-4- O I OT2 + on++ > bn"1"^^^ + 21

Likewise many other systems are found with oxidation potentials

higher than that of iodine-iodide, and they may be reduced by iodide

with the oxidation of the latter to yield an equivalent amount of free

iodine. The iodine so set free may be quantitatively determined, andthus the amount of oxidant liberating it may be computed, by titration

with standard thiosulfate solution. This makes use of the last of the

above reactions; such processes are classified under the heading of

iodometry. Among iodometric determinations are those based upon the

following reactions:

BrOr + 61- + 6H+ -> 3I2 + Br- + 3H2OClOr + 61- + 6H+ -> 3I2 + Cl- + 3H 2OIOr + 51- + 6H+ - 3I2 + 3H2O

Cr2O 7- + 61- + 14H+ - 3I2 + 2Cr+++ + 7H2

2MnO4~ + lOIr + 16H+ - 5I2 + 2Mn++ + 8H2O

2Cu++ + 41- ~> I 2 + 2CuJ 2

H2O 2 + 21- + 211+ -* I, + 2H 2O204

Page 223: quimica inorganica cuantitativa


Determinations based upon such reactions as are listed in tlA first

group above are called direct determinations, because the essential con-

stituent reacts directly with the standard (iodine) solution. In the second

group of reactions we find that the essential constituent reacts to liberate

iodine which, in turn, is titrated with the standard (thiosulfate) solution.

The volume of thiosulfate required, and therefore the calculations in-

volved, are exactly the same as though the essential constituent had

directly oxidized the thiosulfate. The essential constituent may of course

either be a component part of an unknown (copper may, for example,be determined iodometrically) or may be the primary standard used to

standardize our solution (all of the first five in the second group above

are ions the potassium compounds of which often are so employed for

standardizing thiosulfate; copper also may be used). In this indirect

manner sodium thiosulfate may be compared with a great many oxidizing

substances, but it must be remembered that the direct titration of thio-

sulfate against oxidizing agents in general is not feasible. This is because

thiosulfate is quantitatively oxidized to tetrathionate, S^e^, by iodine

alone, in neutral or acid solution. Other oxidants practically without

exception convert thiosulfate more or less incompletely to sulfate. This

also is true even of iodine in weakly alkaline solutions. The proper con-

trol of the pH therefore is of first importance.

The wide use of iodimetry and iodometry in volumetric analysis is

due not only to the high accuracy obtainable by these methods but also

to the availability of an excellent and sensitive indicator, starch. Even

with extremely small concentrations of iodine, starch gives the familiar

deep blue color. The color is attributed to an adsorption complex of

variable composition. The sensitivity of starch as an indicator is greater

in the presence of a small amount of iodide. The sensitivity is impairedsomewhat by higher temperatures arid by the presence of some organic

compounds such as ethyl alcohol. Starch solution should be freshly pre-

pared or else properly preserved. In using the indicator in the titration

of iodine solutions it should not be added until just before the end point

is reached that is, until the deep color of the iodine solution has changedto light amber. If the starch were added at the beginning of the titration

the color of the starch-iodide complex would be practically black arid

there would be little or no warning as the end point was approached.

Furthermore, starch adsorbs an appreciable amount of iodine from a

concentrated solution which is released only slowly to a slight excess of

thiosulfate. A significant error thus would result.

Preparation of Starch Indicator. Make a thin paste by stirring a

little cold water with about 2 g. of soluble starch; then add about 25 ml.

of cold water. Meantime boil a liter of water. Slowly pour the starch

suspension into the hot water while stirring and continue to boil until

Page 224: quimica inorganica cuantitativa


the solution is clear. Cool, add a preservative1

if the solution is to be

stored, and place in a glass-stoppered bottle. As an indicator use about

5 ml. of this solution per 100 ml. of titrated solution.


Preparation of 0.1 N Solution of Iodine. As may be seen in the re-

actions given on p. 204, iodine is reduced to the iodide ion when it

functions as an oxidant; therefore the equivalent weight of iodine is

identical with its atomic weight. Thus to prepare a liter of 0.1 N solu-

tion about 13 g. of iodine is required. Iodine is not very soluble in water

alone, but in an aqueous solution of potassium iodide it dissolves readily

with the formation of the tri-iodide ion, I 3


[ 2 + I- <z I 3


When this solution reacts with a reductant the free iodine acts as the

oxidant. As the iodine is used up the above equilibrium shifts to the left

so that finally all of the iodine, originally present as I 2 and as I 3~, acts in

the oxidi/ing capacity.

Iodine solutions are not stable. They continue to decrease slightly in

normality for years. Standard iodine solutions tend to hydrolyze accord-

ing to the reaction

I 2 + H 2= I- + 211+ + 10-

except in neutral or acidic solutions. In an alkaline solution the action is

accelerated because of the removal of hydrogen ion to form water.

Procedure. Weigh out roughly 13 g. of resublimed iodine from a

stoppered vial by difference (see p. 63), and 40 g. of potassium iodide

free from iodate. Transfer the crystals of both to a 250 ml. beaker andadd 25 ml. or more of water. Stir occasionally until complete solution

has taken place and then pour into a glass-stoppered bottle and dilute

to a liter. The solution should not be exposed to light.

Notes. If the solution is not clear after diluting, it should be filtered throughasbestos in a Gooch crucible.

Test the potassium iodide for iodate by dissolving a crystal in a few milliliters

of water, adding 0.5 ml. of 6 N sulfuric acid and 1 ml. of starch solution. If noblue color develops in 30 seconds the iodide is iodate-free (I0s~ + 5I~ + 6H+= 3I 2 + 3H.>O). Why is iodate objectionable here?

Standardization with Arsenious Oxide. Arsenic trioxide may beobtained well over 99.9 per cent pure As2 3 , and after being dried at 105

1 Various preservatives have been suggested. Naiman, J. Chem. Education, 14, 138(1937), advocates the addition of red mercuric iodide; Nichols, Ind. Eng. Chem., Anal.Ed., 1, 215 (1929), has employed salicylic acid; Platow, Chemist-Analyst, June, 1939,recommends 1 g. of furoic acid per liter of starch solution. Some simply saturate thestarch solution with chloroform.

Page 225: quimica inorganica cuantitativa


it serves as an excellent primary standard for iodine solutions. After being

brought into solution with sodium hydroxide (As2 3 + 20H~ + H 2O =2H 2AsO3-) the oxidation by iodine is effected during the titration accord-

ing to the equation: H 2As03- + I 2 + H2O - H2As0 4

- + 21- + 2H+.

The arsenic changes valence from As+++ to As"f+"


l"+ and since the trioxide

contains two atoms of arsenic it is obvious that the equivalent weight of

arsenic trioxide is one-fourth of its molecular weight.

Procedure. Accurately weigh samples of 0.15 to 0.22 g. of arsenic

trioxide and transfer to 250 ml. Erlenmeyer flasks. Add 10 ml. o( 1 Nsodium hydroxide solution and warm until dissolved. Add 10 ml. (an

excess) of 1 TV hydrochloric acid and then carefully add a solution of

sodium bicarbonate. (2 g. in 50 ml. of water) until the evolution of carbon

dioxide ceases. Next add about 1 g. of sodium bicarbonate to buffer the

solution, dilute to 100 ml., add 5 ml. of starch indicator solution andtitrate with the iodine solution until a faint blue color is obtained. Calcu-

late the normality of the iodine solution. The relative mean deviation

should be not more than 1 part per 1000.

Notes. In the titration of arscnious oxide with iodine the solution must be at a

pH between 5 and 9. Below a pll of 5 the reaction is so decelerated that the end

point becomes rather uncertain. The bicarbonate buffers the solution on the

alkaline side.

If the highest grade of potassium iodide was used in preparing the iodine

solution no blank should be necessary for the standardization, since it contains

no iodate.


Sulfur as sulfide usually is present in small amounts as an impurityin steels. It seldom exceeds 0.05 per cent. The method given here does

not lead to results of highest accuracy but it is rapid and is a more or

less standard procedure for steel analysis. In principle the method con-

sists of treating the steel with acid to form hydrogen sulfide, which is

passed, along with hydrogen and small amounts of other gases which are

formed, into an ammoniacal solution of cadmium chloride which causes

the precipitation of the sulfur as cadmium sulfide. The cadmium sulfide

is acidified and the hydrogen sulfide thus regenerated is treated with

standard iodine solution.

Procedure. Clean a sample of steel turnings with benzene and dry.

Weigh samples of about 5 g. of the steel to the nearest hundredth of a

gram. Place the sample in a 250 ml. Erlenmeyer generator flask. 2 Insert

a two-hole rubber stopper and extend a thistle tube through the stopperalmost to the bottom of the flask. Connect the generator flask to a second

* For apparatus designed especially for this determination see Scott, StandardMethods of Chemical Analysis, D. Van Nostrand Co., New York, 1939, p. 912; also

see Steinmetz, J. Ind. Eng. Chem., 20, 983 (1928).

Page 226: quimica inorganica cuantitativa


250 ml. flask by means of glass tubing. The glass tube should extend

almost to the bottom of the second flask into which is poured 50 ml. of

ammoniacal cadmium chloride solution. This solution is prepared by dis-

solving 5.5 g. of CdCl2.2H2O in 50 ml. of water, adding 120 ml. of con-

centrated ammonium hydroxide and diluting with water to 250 ml. The

flasks should be mounted in such a manner that a burner will go under

the generator flask.

Through the thistle tube pour 75 ml. of 6 N hydrochloric acid and

warm the flask with a small flame until the steel has dissolved com-

pletely; then boil gently for 1 minute, but no longer. Disconnect the

receiving flask but leave the delivery tube, which may now act as a

stirrer. Add 5 ml. of starch solution and 40 ml. of 6 TV hydrochloric acid.

Titrate without delay the solution of hydrogen sulfide thus formed with

standard iodine solution. Calculate the percentage of sulfur in the steel.

Notes. The solution should not be boiled too long since this might cause the

distillation of enough HC1 to neutralize the ammonia and dissolve the cadmium


Problem. A sample of steel weighing 5.00 g. is treated with acid and the hydro-

gen sulfide evolved is passed through ammoniacal cadmium chloride. The cad-

mium sulfide so formed is acidified and the hydrogen sulfide resulting is titrated

with 0.0500 N iodine solution, 6.6 ml. being required. What is the percentage of

sulfur in the steel?

Solution. The number of g. eq. wts. of iodine used is : (0.0500), and the

same number of g. eq. wts. of sulfur must be present. Therefore, since the g. eq.

wt. of sulfur is 3% or 16 (see the equation, p. 204, showing that the electron shift

for the sulfur atom is 2), we have

(0.0066)(0.0500)(16)(100) = g _ ubtUU


Stibnite is an ore consisting largely of antimony sulfide, Sb2S 3 , and

some silica. The following procedure is not suitable for ores containing

appreciable amounts of arsenic or of iron which cause errors unless they

are first separated from the antimony. The principle involved in the

iodimetric determination of antimony is essentially the same as in the

standardization of an iodine solution with arsenic trioxide. Antimony,like arsenic, may be oxidized from a valence of 3+ to 5+ and the essential

reaction of the titration may be seen on p. 204. Actually, because tar-

taric acid is added during the course of the analysis, the reaction is:

H(SbO)C 4H 4 6 + I 2 + H2= H(Sb0 2)C 4H4O6 + 2I~ + 2H+. The ti-

tration is quantitatively complete only if the solution is neutral or slightly

alkaline; if the pH is too low iodide reduces antimonate, while if too high

the reaction between iodine and water (I 2 + H2= I~ + 10"" + 2H+)

Page 227: quimica inorganica cuantitativa


is driven to the right because the hydroxyl ions remove the hydrogenions. In the presence of sodium bicarbonate the reaction proceeds satis-

factorily. Provision also must be made to prevent loss of antimonychloride by volatilization when the ore is being dissolved in hydrochloricacid. This is accomplished by careful control of the temperature and bythe addition of potassium chloride which increases t*he chloride ion con-

centration, and this, in turn, increases the concentration of the complexion, SbCl4~.

It is also necessary when the solution is diluted to prevent the pre-

cipitation of basic antimony chloride, SbOCl, which may be formed

by hydrolysis. Hydrolysis may be prevented, however, by controlling

the acidity and by forming a complex antimony tartrate as alreadymentioned.

Procedure. Accurately weigh out samples of the finely divided, dryore of about 0.5 to 1 g. each. Transfer to 400 ml. beakers and add 20 ml.

of concentrated hydrochloric acid and 0.3 to 0.4 g. of potassium chloride.

Cover with a watch glass and heat on a steam bath for 15 minutes or

until the residue of silica is white. Add 3 g. of tartaric acid and heat on

the steam bath for 10 minutes longer. Slowly add, with constant stirring,

100 ml. of cold water. Should a red precipitate of antimony sulfide appearat this point it indicates that hydrogen sulfide was not completely ex-

pelled. In that case, stop the addition of water, heat the solution againover the steam bath until the precipitate dissolves; then cool and add

the remaining part of the 100 ml. of cold water. Should a white precipi-

tate of basic antimony chloride appear when the solution is diluted it

will be necessary to start over from the beginning, since it is practically

impossible to bring the basic salt back into solution. Add 4 drops of

methyl red indicator and 6 TV sodium hydroxide solution until the solu-

tion has just turned yellow; then add 6 N hydrochloric acid, drop by

drop, until the solution is barely acid, i.e., faintly pink or orange. After

this add 3 g. of sodium bicarbonate in 100 ml. of water, and 5 ml. of

starch solution. Then titrate with standard iodine solution. The approachto the end point should be made with utmost care since the reaction

between trivalent antimony and iodine is relatively slow and the end

point might easily be overrun. Calculate the percentage of antimony in

the ore. The relative mean deviation might run as high as 3 or 4 parts

per 1000.


This substance, KSbC^Oe.Jl^O, is used in medicine to cause

emesis. Since one molecule of the compound contains one atom of tri-

valent antimony the analysis may be carried out essentially as outlined

for atibnite.

Page 228: quimica inorganica cuantitativa


Procedure. Weigh accurately samples of 0.5 to 0.6 g. Add 100 ml. of

water to dissolve, then about 1 g. of tartaric acid and 2 g. of sodium

bicarbonate. Put in 5 ml. of starch solution and titrate with standard

iodine solution. Calculate the percentage purity of the salt. The relative

mean deviation should not exceed 2 parts per 1000.


Preparation and Standardization of 0.1 N Sodium Thiosulfate.

Any substance which meets the criteria of primary standards in general

(p. 78) and which quantitatively oxidizes iodides to iodine may be used

to standardize a solution of sodium thiosulfate. The substances most

commonly used are the potassium salts of bromate, iodate and dichro-

mate. The bi-iodate also often is used. Metallic copper may be employed

since cupric salts react with iodide in weakly acid solution to liberate

iodine, but this method is recommended only when the thiosulfate solu-

tion is to be used for the determination of copper. Whatever standard is

selected, the accurately weighed substance reacts in an acid solution of

potassium iodide to set free a definite quantity of iodine which then is

titrated with the thiosulfate solution the normality of which is sought:

2S 2O35!=t + 12

- S406~ + 2I~. If the bromate or iodate is used as the

standard it is evident that no colored product is formed save the iodine

itself, with the result that the titration with thiosulfate, using starch as

the indicator, gives a color change at the end point from blue to colorless.

On the dther hand, if potassium dichromate is used as the standard, the

greenish chromic ion is formed in the reaction with the iodide ion. Thus

when the solution is titrated with thiosulfate the color change at the end

point is from blue to light green. This is sometimes mentioned as a dis-

advantage to the use of dichromate as the standard. As a matter of fact,

unless one is partially color-blind this presents not the slightest difficulty,

and because of other advantages (e.g., the equivalent weight of potassium

dichromate is greater than that of any of the other three) it is preferred

by many analysts over other primary standards.

Before taking up the actual procedure for making up and standardizing a

solution of sodium thiosulfate let us examine the reactions between iodides and

bromates and between iodides and iodates. The equation for the former reaction


(1) Br03~ + 61- + 6H+ - 3I 2 + Br- + 3H2

The iodide ions lose one electron each to become iodine atoms, or molecules;

at the same time the brtimate ion is reduced to bromide ion, and this involves a

gain of six electrons. It is obvious then that the equivalent weight of potassiumbromate is its molecular weight divided by 6, or 167.01/6 27.84,

Page 229: quimica inorganica cuantitativa


Consider now the corresponding reaction between iodate and iodide ions. The

equation is sometimes written (in a manner analogous to the bromate reaction)

(2) I0 3- + 61- + 6H+ - 3I 2 + I- + 3H 2

However, here we have iodide ion appearing on both sides of the equation; it

obviously may be condensed to read

(3) I0 3- + 51- + 6H+ - 3I 2 + 3H 2

According to equation (2) the equivalent weight of potassium iodate would be

one-sixth of its molecular weight, but noting in equation (3) that 5 iodide ions are

oxidized, we conclude that the equivalent weight of the iodate is one-fifth of its

molecular weight.

Though we might expect bromate and iodate to act alike there is no anomalyhere. It ought to be evident that if the same number of moles of potassium iodate

and potassium bromate were weighed out, acidified and titrated directly with

potassium iodide solution (electrometrically, for since the iodine color appears

immediately the color indicator, starch, could not be used to mark the end point),

% more iodide solution would be necessary in the bromate titration than in the

iodate titration. This is mathematically evinced by equations (1) and (3). Or

sometimes it is reasoned as follows, though such reasoning is arguing in circles:" The iodate ion is reduced to iodide ion as equation (2) implies, but it could never

remain in solution as such, since more iodate would oxidize it, the iodide, to iodine.

This means that just that much less iodide would have to be added from the buret

to satisfy the needs of the given weight of iodate." This reasoning automatically

leads to the conclusion that equation (3) is the correct representation and not

equation (2). Though the conclusion is right the fallacy of such reasoning is that

it assumes that iodine in the iodate would pass from a valence of 5 + to a valence

of 1- in iodide, and then return to a valence of zero in molecular iodine. It is only

logical to say that the iodate ion is reduced directly to 1 while, simultaneously,

iodide ion (from the buret) is oxidized to 1.

Of course the chief objection to writing the equation as expressed in (2) is

that it leads to the erroneous conclusion that the equivalent weight of potassium

iodate, when titrated with potassium iodide, is one-sixth of the molecular weight,

whereas actually it is one-fifth. (Note, however, that equation (1) shows that if

potassium bromate is titrated with potassium iodide, the equivalent weight of the

bromate is one-sixth of its molecular weight. The reduction of bromate by iodide

could not possibly stop at the elementary bromine state since bromine is reduced

by iodide to bromide.)

Nothing that has been said above belies the fact that, used as a primary stand-

ard in the standardization of sodium thiosulfate solution, the equivalent weight

of potassium iodate is one-sixth of its molecular weight, just as is the case with

potassium bromate. Consider again equations (1) and (3); in both instances one

mole of the halate liberates three moles of iodine (3I 2). Therefore, when the

iodine is titrated with thiosulfate

the same amount of thiosulfate is required. In brief, a mole of either bromate or

iodate gives rise to that quantity of iodine which oxidizes six moles of thiosulfate.

Each molecule of thiosulfate when oxidized loses one electron. It follows then that

the equivalent weight of both potassium bromate and potassium iodate, when

Page 230: quimica inorganica cuantitativa


used in iodometry, is one-sixth the molecular weights. The fact that potassium

iodate has an equivalent w:ight which sometimes is one-fifth and sometimes

one-sixth of its molecular weight should remind us of a statement made earlier

(p. 153) that "it is imperative that one consider the particular reaction which the

reagent is undergoing . . . before deciding what is the equivalent weight."

Sodium thiosulfate solution is very difficult to prepare in such a

manner that it will maintain a constant normality. Often the concentra-

tion will vary significantly even within a period of one or two days. The

change in normality may be either positive or negative. Oxidation by

atmospheric oxygen may occur:

3- + O 2 -> 2SOr + 2S

which may account for a decrease in normality. The bacterium, Thio-

bacillus thioparus, which is said to be ordinarily present in air,3slowly

decomposes the solution. The gain in normality may be explained as a

result of acidity, possibly due to the presence of carbon dioxide:

Thus for each thiosulfate ion lost there is a gain of one sulfite ion, and

since the latter ion has twice the reducing capacity of the thiosulfate ion

(gaining two electrons when oxidi/ed to S04"

as compared with one

electron gained by S 2O 3~

in iodometry), the not result is an increase in

the normality of the solution as a reductant. To overcome these diffi-

culties a great many suggestions have been made. Boiling the water in

which the thiosulfate is to be dissolved destroys the bacteria and also

expels carbon dioxide. Of course the storage bottle and its stopper should

be sterile. Sodium carbonate, borax, sodium furoate and chloroform have

been used as preservatives.

Procedure. Weigh out approximately 25 g. of sodium thiosulfate

pentahydrate and transfer to a liter Pyrex bottle which has been sterilized

with hot water. Boil a liter of distilled water and pour while quite hot

into the bottle. Plug the mouth of the bottle with sterile cotton while

the solution cools, then add as a preservative either 2 g. of borax, 0.5 g.

of sodium carbonate, 1 g. of sodium furoate or several drops of chloroform.

Stopper the bottle and shake.

Standardization with Potassium Bichromate. The reaction

between the dichrornate and iodide ions in an acid medium

Cr2O 7- + 61- + 1411+ -> 3I 2 + 2Cr+++ + 7H2O

requires a rather high acidity if it is to proceed rapidly. The standardiza-

tion can be carried out at a lower acidity if air is expelled from the flask

3 Kassner and Kassner, /net. Eng. Cftem., Anal Ed., 12, 655 (1940).

Page 231: quimica inorganica cuantitativa


with carbon dioxide, thus avoiding the danger of oxidation of iodide byair, which may happen in strongly acid solutions. 4

Procedure. Accurately weigh out samples of about 0.2 g. (confirm

this value) of pure potassium dichromate which has been well dried at

140, and dissolve in 40 to 50 ml. of water. Add 8 ml. of concentrated

hydrochloric acid. Meantime, fill the buret with the thiosulfate solution

which is to be titrated and set the meniscus on the zero mark. Prepare the

starch indicator solution if this has not already been done. Next quickly

dissolve about 3 g. of pure potassium iodide in 10 or 15 ml. of water,

add it immediately to the acidified dichromate solution, dilute to about

150 ml. and stir. Set in a dark place for 3 minutes only and then titrate

with thiosulfate, stirring constantly, until the dark amber color gives

way to an orange-green. Then add 5 ml. of starch indicator and slowly

continue the titration until the blue color suddenly is replaced by a

clear, light green. Run a blank using the same quantities of all reagents

except potassium dichromate. Calculate the normality of the thiosulfate

solution. The relative mean deviation should not exceed 1 part per 1000.

Notes. The potassium iodide solution should not be prepared until everythingelse is in readiness because the solution will throw out iodine if allowed to stand

in the light and exposed to air even for a few minutes. After bringing the solutions

of potassium iodide and acidified potassium dichromate together they should be

placed in the dark for 3 minutes to allow time for them to react quantitatively.Too long a period, though, may result in loss of iodine through volatilization or a

gain in iodine by oxidation (4I~ + 2 + 411+ > 21 2 + 2H 2O), either of which

would introduce an error into the titration which follows.

The solution should be stirred or swirled constantly while titrating in order ton+

prevent the reaction, S aOs" =- SO3" + S, due to a local excess of thiosulfate in

the acid solution.

A blank run is necessary if the potassium iodide crystals contain small amountsof the oxidant, potassium iodate.

Standardization with Potassium Bromate, Iodate or Bi-

iodate. Because the equivalent weights of these three standards are not

widely different, being respectively 27.81, 35.67 and 32.50 (confirm),

about the same amount of any may be used in the standardization of

thiosulfate. The manner in which they react with potassium iodide has

already been explained (p. 210).

Procedure. Accurately weigh out samples of about 0.10 to 0.12 g.

of pure, dry potassium bromate, potassium iodate or potassium bi-iodate

and dissolve in 40 to 50 ml. of water. Add 8 ml. of concentrated hydro-chloric acid. Fill the buret with the thiosulfate solution which is to be

standardized and set the meniscus on the zero mark. Prepare the starch

4 Bray and Miller, J. Am. Chem. Soc.. 46, 2204 (1924); Popoffand Whitman, ibid.,

47, 2259 (1925); Hahn, ibid., 57, 614 (1935).

Page 232: quimica inorganica cuantitativa


indicator solution if this has not already been done. Next quickly dis-

solve about 3 g. of pure potassium iodide in 10 or 15 ml. of water, add it

immediately to the acidified solution of the bromate (or iodate or bi-

iodate), dilute to about 150 ml. and stir. Set aside in a dark place for

about 3 minutes and then titrate with the thiosulfate until the solution

is light amber in color. Add 5 ml. of starch indicator and slowly continue

the titration until the blue color gives way to colorless. Run a blank

using the same quantities of all reagents except the oxidant, which is

omitted. Calculate the normality of the thiosulfate solution. The relative

mean deviation should not exceed 1 part per 1000.

Notes. Read the notes following the procedure for standardizing the solution

with potassium dichromate.

Standardization with Copper, Copper is recommended as a pri-

mary standard for thiosulfate only when the latter is to be used for the

determination of copper. Under proper conditions copper ions oxidize

iodide ions quantitatively :

and the liberated iodine may be titrated with thiosulfate. A weighed

quantity of copper is dissolved in nitric acid. The excess nitric acid and

oxides of nitrogen are expelled by evaporation with sulfuric acid. Then

the pH is adjusted to about 3.7 and the solution titrated.

The oxidation-reduction represented by the above equation takes

place despite the fact that the oxidation potential of the iodine-iodide

system is above that of the cupric-cuprous system (see Table 17, p. 171).

This is accomplished by keeping the concentration of cuprous ion at a

very small value since cuprous iodide is precipitated, and by making the

concentration of the iodide ion quite large through adding a considerable

excess of potassium iodide.

The end point in this titration is not difficult to recognize but is

somewhat unusual in that the precipitate of cuprous iodide is present.

This precipitate adsorbs a quantity of iodine, insufficient to affect ap-

preciably the results of the standardization, but sufficient to change the

normally white cuprous iodide to a cream or ivory color. The color changeat the end point, therefore, is not from blue or lavender to white but

from lavender to ivory.

Procedure. Accurately weigh samples of clean, bright copper wire of

about 0.2 to 0.25 g. each. Place the copper in a casserole and add 5 ml.

of concentrated nitric acid and allow to dissolve. If necessary add another

5 ml. of nitric acid. When dissolved add 2 ml. of concentrated sulfuric

acid and evaporate carefully over a very low flame until sulfuric acid

fumes appear. Avoid spattering. Cool and very cautiously add 25 ml. of

Page 233: quimica inorganica cuantitativa


water, and then slowly add 6 N ammonium hydroxide until the solution

barely attains a deep blue color. Next add 20 ml. of 6 N acetic acid. Thesolution should now have a pH of about 3.7. Dissolve 3 g. of pure po-tassium iodide in 10 or 15 ml. of water and add at once to the coppersolution. Titrate the liberated iodine with the thiosulfate solution until

the muddy brown color has almost faded; then add 5 ml. of starch solu-

tion and complete the titration. Calculate the normality of the thio-

sulfate solution. The relative mean deviation should not exceed 2 parts

per 1000.


The analysis of a substance which contains no interfering elements

may be carried out by essentially the same procedure as has been givenfor the standardization of thiosulfate with metallic copper. But manyores of copper contain, besides silica, such metallic impurities as alumi-

num, iron, silver, arsenic and antimony. Silver and aluminum cause no

trouble in the analysis for copper, but ferrous iron, trivalent arsenic and

antimony do interfere because they act as rcductants. Ferric iron cannot

be tolerated since it acts as an oxidant. Although pentavalent arsenic

and antimony react with the iodide ion, they do so only slowly under

the proper condition of acidity (pH > 3.5), and their presence causes noserious error. The ore is brought into solution with nitric acid or with

aqua regia and the oxides of nitrogen are expelled by evaporation with

sulfuric acid, for they would oxidize iodide if allowed to remain in solu-

tion. Any arsenic or antimony is oxidized to the harmless pentavalentstate with bromine and the excess bromine is eliminated by boiling. The

acidity is fixed at a pH around 3.7 and the iron, which was oxidized to

the ferric state by bromine, is decreased to a negligible concentration of

ferric ion by converting it into the complex FeFe=

ion with ammoniumbifluoride. The solution then may be treated with excess potassium iodide

and the iodine liberated by the cupric ion titrated with the standard

thiosulfate solution.

Procedure. Accurately weigh samples of the finely divided, dry ore

of about 0.5 g. or more and transfer to 250 ml. beakers. Add 10 ml. of

concentrated nitric acid and heat over a very low flame until all that

remains undissolved is a white siliceous residue. If solution of the copperis incomplete, add 5 ml. of concentrated nitric acid and 5 ml. of concen-

trated hydrochloric acid, heat the solution and cool. Add 5 ml. of concen-

trated sulfuric acid. Heat cautiously over a hot plate until sulfuric acid

fumes are evolved. Cool and very carefully add 25 ml. of water. If the

residue is white it need not be filtered; if dark, filter and wash residue

and filter paper thoroughly.

When the copper is in solution add 5 ml. of saturated bromine water

Page 234: quimica inorganica cuantitativa


and boil gently for several minutes to expel the excess bromine. (Test

vapors for bromine with starch-iodide paper prepared by dipping a piece

of filter paper in a solution of starch and potassium iodide.) Cool the

solution, add 6 N ammonium hydroxide until a faint permanent pre-

cipitate of ferric hydroxide has been obtained, or in the absence of iron.

until, after stirring, a slight odor of ammonia persists. Then add 20 ml.

of 6 N acetic acid. Test the solution with hydrion paper by touching the

paper to the stirring rod in the beaker. If the pH is riot about 3.7, adjust

by adding a few drops of acetic acid or of ammonium hydroxide. Add

2 g. of ammonium bifluoride. Stir until dissolved and then add a freshly

prepared solution of 3 g. of potassium iodide in 10 or 15 ml % of water.

Titrate the liberated iodine at once with the standard thiosulfate solu-

tion, using starch indicator after the muddy brown color has changed to

very light brown. Continue adding thiosulfate drop by drop until the

lavender color changes to ivory. If desired, 2 g. of ammonium thiocyariate

may be added just before the end point; this reduces the amount of iodine

adsorbed by the precipitate (see discussion under Standardization with

Copper, p. 214) and causes a sharper end point.5 Run a blank and make

the necessary correction. Calculate the percentage of copper in the ore.

The relative mean deviation should not exceed 3 parts per 1000.

Notes. After the end point is reached no blue color should return for at least

10 minutes. An almost immediate return of color indicates too much ammonium

hydroxide or too little bifluoride, or oxides of nitrogen.


Bleaching powder has the formula Ca(OCl)Cl. The compound is

capable of reacting with acids to liberate chlorine.

CaOCl, + 211+ -> Ca++ + C12 + H 2O

As a bleaching agent or a disinfectant the value of bleaching powder

depends upon the amount of chlorine which it makes "available." If

bleaching powder is allowed to react with an excess of potassium iodide

it liberates free iodine,

OC1- + 21- + 2H+ -> I 2 + Cl- + H2

which then may be titrated with standard thiosulfate solution.

Procedure. Accurately weigh out to the third decimal place about 5

g. of commercial bleaching powder and grind it in a mortar, adding small

quantities of water and grinding thoroughly after each addition until the

powder becomes a smooth paste. Wash the contents of the mortar quanti-

tatively into a 500 ml. volumetric flask and dilute to the mark with water.

6 Foote and Vance, J. Am. Ghent. Soc., 57, 845 (1935); hid. Eng. Chem., Anal. Ed.,

8, 119 (1936).

Page 235: quimica inorganica cuantitativa


Shake the suspension well and transfer 50 ml. portions by means of a

pipet to 250 ml. conical flasks, observing the precaution that the liquor

withdrawn contains in each case its proportion of suspended matter. Di-

lute the 50 ml. portions to 100 ml., add a freshly prepared solution of 3 g.

of potassium iodide in 10 ml. of water, and 10 ml. of 6 TV sulfuric acid.

Titrate the liberated iodine with standard thiosulfate solution, adding5 ml. of starch indicator solution just before the end point. Calculate

the percentage of available chlorine. Duplicate results should agree within

a relative mean deviation of 1 part per 1000 if the volumetric flask is

well shaken before removing the aliquot portions with the pipet.


The reaction of hydrogen peroxide and potassium iodide in the pres-

ence of acid proceeds according to the equation,

H2 2 + 21- + 2H+ -> I 2 + 2H 2O

and though it proceeds slowly it may be catalyzed by the molybdate ion.

The iodine liberated is titrated with standard thiosulfate solution.

Procedure. A sample of commercial hydrogen peroxide is diluted in a

volumetric flask in the manner described in the procedure for analysis

by the permanganate method, p. 199. Pipet aliquot portions of 25 ml.

into 125 ml. conical flasks, acidify with 8 ml. of 6 TV sulfuric acid, add a

freshly prepared solution of 3 g. of potassium iodide dissolved in 10 ml.

of water and 3 drops of neutral 1 N ammonium molybdate solution.

Titrate the liberated iodine with standard thiosulfate solution, adding5 ml. of starch indicator solution just before the end point is reached.

Prove from the equations involved that 1 ml. of tenth-normal thiosulfate

solution is equivalent to 0.001701 g. of hydrogen peroxide. Calculate the

percentage of H2O 2 in the sample. The deviation from the mean may be

as high as 4 parts per 1000.


1. A solution contains 6.355 g. of KHC 2O 4.H 2C 2O 4.2H 2O per 100.0 ml. (a) Whatis its molarity ? (b) What is its normality as an acidP (c) What is its normalityas a reducing agent?

Answer: (a) 0.2500; (b) 0.7500; (c) 1.000.

2. Exactly 40.00 ml. of a KHC 2 4 solution is equivalent to 25.00 ml. of a potas-sium hydroxide solution which contains 11.22 g. of KOH per liter. Howmany milliliters of acidified potassium permanganate solution containing10.54 g. ofKMn04 per liter will react with 30.00 ml. of the oxalate solution?

Answer: 22.49 ml.

3. A solution contains 9.488 g. of Ce(S0 4)2.2(NH 4) 2S04.2H 2 per 50.00 ml. If a

stannous tin solution is oxidized by 35.36 ml. of the eerie solution, how manygrams of tin are present ?

Answer: 0.6296 g.

Page 236: quimica inorganica cuantitativa


4. Fifty ml. of a solution of potassium dichromate containing 25.00 g. of K 2Cr2O 7

per liter is acidified with sulfuric acid and poured into 140.0 ml. of ferrous

ammonium sulfate hexahydrate containing 100.0 g. ofFeSO^CNH^SO^I^Oper liter. Does the resulting solution contain excess oxidizing or reducing

agent? How many milliliters of 0.4000 TV oxidizing or reducing solution will

be required in order that no excess of oxidant or reductant be present?

Answer: 25.53 ml. of oxidant.

5. Calculate the normality of a solution of potassium permanganate of which

38.46 ml. is necessary to react with 0.2698 g. of sodium oxalate.

Answer: 0.1047 N.

6. The potassium permanganate solution of problem 5 has what arsenious

oxide value; i.e., what weight of As20j is equivalent to 1.000 ml. of the

permanganate?Answer: 0.005178 g.

7. Calculate the normality of a eerie ammonium sulfate solution of which 33.33

ml. is found to be equivalent to 0.1948 g. of pure iron wire.

8. What is the normality of a sodium thiosulfate solution of which 40.42 ml.

reacts with the iodine set free from excess potassium iodide by 0.1177 g. of

potassium bromate?

9. The sodium thiosulfate of problem 8 above has what potassium bi-iodate


Answer: 0.003399 g.

10. Calcium oxakite is precipitated from a 0.3608 g. sample of limestone con-

taining 41.75 per cent CaO. The precipitate is treated with excess sulfuric

acid and titrated with a potassium permanganate solution, 42.52 ml. being

required. Calculate the molarity of the permanganate solution.

Answer: 0.02526 M.

11. Approximately 5 ml. of a hydrogen peroxide solution is found to weigh 5.162

g. It is titrated with a 0.1050 N potassium permanganate solution. The buret

reading is 40.02 rnl. Calculate the percentage of H 2 2 in the sample.

Answer: 1.39%.

12. A potassium permanganate solution is of such concentration that 1.000 ml.

of it is equivalent to 0.002900 g. of potassium bromate. One ml. of a second

permanganate solution is equivalent to 0.01000 g. of arsenious oxide. Howmany milliliters of the first permanganate solution must be added to 100.0

ml. of the second permanganate solution in order that the resulting normalitywill be 0.1500?

Answer: 114.0 ml.

13. A certain volume of a potassium dichromate solution will oxidize a weight of

potassium tetroxalate, KHC 2Oi.H 2C 2 4.2H 20, which requires four-tenths of

that volume of 0.2500 N sodium hydroxide for neutralization. If this same

potassium dichromate solution is used to analyze an iron ore, and 40.20 ml.

is necessary in the titration of the ferrous solution from a 1.5000 g. sampleof the ore, what is the percentage of iron in terms of FezOs ?

Answer: 28.53%.

14. A sample of pyrolusite,. impure Mn0 2, weighing 0.6022 g. is treated with anexcess of hydrochloric acid and the chlorine which is evolved is passed into a

potassium iodide solution. The iodine thus liberated is titrated with a sodium

Page 237: quimica inorganica cuantitativa


thiosulfate solution (of which 1 1. is equivalent to 12.71 g. of copper), 41.56 ml.

being used. Calculate the percentage of Mn0 2 in the pyrolusite.

Answer: 60.01%.

15. The Mn0 2 in a sample of pyrolusite, crude Mn0 2 , is equivalent to 22.02 ml.

of sodium thiosulfate (1.000 ml. of which reduces the iodine liberated from

potassium iodide by 0.005025 g. of potassium dichromate). The sample is

treated with a weighed excess of sodium oxalate, in the presence of sulfuric

acid, and titrated with 0.1009 N potassium permanganate solution. If 32.25

ml. of the permanganate solution was used, what 'weight of sodium oxalate

must have been added?

Answer: 0.3692 g.

16. Calculate the weight of potassium iodate necessary to liberate that amount of

iodine from potassium iodide which will react with 45.67 ml. of a sodium thio-

sulfate solution, 38.24 ml. of which is equivalent to 40.20 ml. of an iodine

solution, if 45.67 ml. of the iodine solution is equivalent to 0.2420 g. of anti-

mony trioxide.

Answer: 0.1245 g.

17. A 0.5000 g. sample of chromite ore (crude FeO.Cr 2 3) is fused with sodium

peroxide, acidified, and the resulting chromate solution is divided into two

equal parts. One portion is analyzed for chromium by adding to it 40.00 ml.

(an excess) of freshly prepared, pure ferrous sulfate solution containing 5.560

g. of FeS0 4.7H 2 per 100.0 ml.; 22.42 ml. of potassium dichromate solution

(1.000 ml. is equivalent to 0.005000 g. of pure iron) is required to titrate the

excess ferrous sulfate. (a) Calculate the percentage of chromium in the ore.

(b) The other portion is added to excess potassium iodide. How many milli-

liters of sodium thiosulfate will be necessary to titrate the liberated iodine if

1.000 ml. of thiosulfate is equivalent to 0.01222 g. of Cu?Answer: (a) 41.55%; (b) 31.16 ml.

18. A solution contains only sulfuric acid, chromic acid and water. Exactly 1 g.

is diluted and titrated with 0.1100 N ferrous sulfate solution, 38.50 ml. being

required. Another 1 g. sample is titrated with 0.1525 N sodium hydroxidesolution for total acidity; 48.20 ml. of the base is necessary. Calculate the

percentage of each acid present.

Answer: 16.66% H 2Cr0 4 ; 22.20% H 2S04 .

19. Thirty ml. of potassium permanganate solution which has been standardized

against sodium oxalate and found to be 0.1000 TV is added. to an excess of

manganous sulfate with which it reacts thus:

2KMnO 4 + 3MnS04 + 2H 2= 5Mn0 2 + 2KHSO 4 + H 2S0 4

Calculate the number of milliliters of 0.05000 N sodium hydroxide solution

which will be needed to neutralize completely the solution resulting after the

above reaction.

Answer: 24.00 ml.

20. A manganese ore weighing 0.1856 g. was brought into solution with the man-

ganese in the Mn++ state. It was titrated in neutral solution with standard

potassium permanganate solution:

2MnOr + 2H 2 5MnO, + 4H+

Page 238: quimica inorganica cuantitativa


The titration required 48.20 ml. of permanganate. The permanganate, as

standardized against sodium oxalate in acid solution, has a normality of

0.1111. Calculate the percentage of manganese in the ore.

Answer: 47.6%.

21. A sample weighing 0.2500 g. and consisting entirely of Fe and FesCh is dis-

solved, reduced completely to ferrous iron and titrated with 0.1000 N eerie

ammonium sulfate solution, 40.00 ml. being required. Calculate the percent-

ages of Fe and Fe 3 4 in the sample.Answer: 61.5% Fe; 38.5% Fe3 4 .

22. In calculating the normality of an oxidizing solution which w,as standardized

against oxalate, an analyst used the equivalent weight of potassium oxalate

and obtained 0.0782 as the normality. He then discovered that actuallysodium oxalate had been used. What is the correct normality of the oxidizing

solution ?

Answer: 0.0970 N.

23. How many grams of sulfur dioxide will be needed to destroy the odor caused

by hydrogen sulfide in one cubic meter of water:

S0 2 + 2H 2S = 38 + 2H 2

if it is found that 1 1. of the water, when treated with 40.00 ml. (an excess) of

0.1000 N iodine solution, requires 19.84 ml. of 0.1092 N sodium thiosulfate

solution to titrate the excess iodine?

Answer: 29.4 g.

24. The sulfur occurring as sulfide in a 5.00 g. sample of steel is determined as out-

lined on p. 207, except that the hydrogen sulfide is passed into an excess of

standard iodine solution and the iodine remaining after reaction with the

hydrogen sulfide is titrated with standard sodium thiosulfate solution. Thevolume of iodine solution receiving the hydrogen sulfide was 20.00 ml. andthe thiosulfate necessary for the titration was 18.06 ml. One ml. of the iodine

solution is equivalent to 0.990 ml. of the thiosulfate solution and 1.000 ml. of

the latter is equivalent to 0.00536 g. of Cu 20. Calculate the percentage of

sulfur in the steel.

Answer: 0.042%.

25. A sample of potassium iodide of commercial grade weighing 0.3893 g. is dis-

solved in water and added to an acidified solution of 0.1141 g. of pure potas-sium bromate. The solution is boiled gently until all of the liberated iodine is

expelled. It then is cooled and an excess of reagent potassium iodide is added,after which it is titrated with 18.02 ml. of sodium thiosulfate solution of which1.000 ml. is equivalent to 0.003567 g. of potassium iodate. Calculate the

purity of the commercial potassium iodide.

Answer: 9B.Q%.

Page 239: quimica inorganica cuantitativa

Chapter 13



[RAVIMETRIC analysis includes those determinations in which the con-

stituent sought is isolated and weighed. The constituent may be sepa-rated from other components in the original sample in the form of the

constituent itself or as one of its compounds. The electrodeposition of

copper on the cathode during the electrolysis of one of its soluble com-

pounds illustrates the former. The latter type is illustrated by the de-

termination of carbon dioxide from a carbonate sample in which the gasis trapped after evolution in an absorbent such as sodium hydroxide, and

weighed. By far the most common type of gravimetric analysis, however,is that in which the essential constituent is converted into a difficultly

soluble compound which is filtered, washed, dried and weighed. In this

last case it is imperative that the compound which is precipitated be

thrown out of solution quantitatively, which is to say it must be practi-

cally insoluble. It should be a substance which presents no particular

difficulty in filtering and washing. It must be a compound of definite

composition or must be converted into a compound of definite compo-sition in a simple manner, as by ignition, in order that its weight mayhave a known mathematical relationship to the original sample.


The formation of a precipitate consists of throwing a substance out of

solution and into the solid state. A true solid is a substance which has as-

sumed a crystalline form in which the ions of the compound are arrangedin some symmetrical fashion about an imaginary three-dimensional trellis

known as the crystal lattice. There are strong forces which keep the ions

in their orderly arrangement in the lattice, forces which must be over-

come in order to disrupt the architecture of the crystal. This may be

done by a sufficiently high temperature (melting or sublimation) or bydissolving in some solvent. In the latter case obviously there is an at-

traction between the ions of the crystal and those in the solution which

is greater than the intracrystalline forces. The extent of solubility of

a given compound in a given solvent must depend upon the relative

magnitudes of the two forces. It is not surprising, then, that solubility


Page 240: quimica inorganica cuantitativa


should vary so widely for different compounds in a given solvent or for

the same compound in different solvents. Nor should one expect solu-

bility to be independent of temperature since the kinetic energy of mole-

cules and ions varies with a change in temperature. A saturated solution,

being one in which the solution is in equilibrium with the solute, is a

solution in which the attractive forces, between ions within the crystal

and between ions of the crystal and of the solution, are equal. When this

equality has been established the rate at which ions are leaving the

crystal lattice is equal to the rate of deposition of ions from the solution

upon the lattice.

Solubility Product. The equilibrium between a salt and its saturated

solution may be studied from the standpoint of the Law of Mass Action.

When the solid, BA, is dissolved in water, sooner or later the solution

becomes saturated and the equilibrium may be expressed by the equation

(1) BA 4=> B+ + A-(solid)


It will be recalled that the Law of Mass Action states that for equi-

libria in general the equilibrium constant is equal to the product of the

concentrations of the ions or molecules on the one side of the equation

divided by the product of the concentrations on the other side, all raised

to a power represented by the coefficients of the ions or molecules. Thus

the mass law expression for the above reaction is simply

m [B+KA-] _(2) -IBXT

Since solid is present and the solution is saturated the value of [BA] is

constant and the denominator may be combined with K, yielding

(3) [B+][A-] = #.,.

where KB .V .is known as the solubility product.

2 The solubility product

of a salt of a two-one valence type, for example B 2A, is, of course

(4) [B+]*[A-] = #.,.-

1 The solubility of any salt is measured by the concentration of the

'saturated solution. Assuming that a salt is completely ionized, it is a

simple matter to calculate the value of KB.V .from the solubility or vice

versa. A salt yielding two univalent ions will have a solubility S, equal

1 As already implied, the compound BA in the solid form is comprised of B4"

and A~"

ions arranged in a definite pattern, but to avoid confusion in discussing the solid-

solution equilibrium the solid state may be designated as BA.2Strictly speaking, the activities of the ions should be used instead of the concen-

trations. See p. 105 for a brief discussion of the activity concept.

Page 241: quimica inorganica cuantitativa


to the concentration of either ion; for example, in the case of silver


S = [AgCl] = [As*] = [C1-]

where the concentrations are expressed in moles per liter. If the solubility

of silver chloride is measured and at a certain temperature is found

to be about 0.0014 g. per liter, then it has dissolved to the extent of

0.0014/143.3 = 10~"6 moles per liter. Since one molecule gives one each

of the ions, the complete dissociation of the salt will establish a value of

10- 6 for both [Ag+] and [Cl~] so that

#.p. = [Ag+][Cl-] = [10-*]2 =

Conversely, knowing the value of KB.. to be 10~ 10, the solubility would

be calculated thus:

S = VlO~ 10 = 10-5 moles per liter, or

S = (10-6) (143.3) ^ 0.0014 g. per liter

The solubility of a salt of the B 2A type, like Ag 2Cr04, usually is

taken as equal to the concentration of the bivalent ion, since each

molecule upon ionizing yields one bivalent ion whereas it gives two of

the univalent ions. Therefore it follows that the value of [B+] must be

twice that of [A~], and if we let [A~] = X then equation (4) becomes

(2X)2(X) ==#..,.


(5) 4X' = K.,.p.

from which, if #8 .p. is known, the solubility is calculable. To illustrate:

Knv .for silver chromate has a value of 2.0 X 10~~ 12 . Therefore

4X3 = 2.0 X 10- 12

x - . , x

The solubility of silver chromate thus is 7.9 X 10" 5 moles per liter, or

(7.9 X 10- 5)(332) = 0.026 g. per liter for the temperature taken.

Because the solubility product is based upon equilibrium conditions

between solute and solution that is, a saturated solution it follows

that if the concentrations of the ions of a given salt are such that their

product (raised to the proper exponential power) exceeds the value of

KB.v.j precipitation of the salt must take place until the ion product is

decreased to the /C,.p. value. If the ion product gives a value less than

that of KB .V ,the solution will be unsaturated, while a solution containing

the ions in such concentrations that their product just equals the value

Page 242: quimica inorganica cuantitativa


of / 8.p. must be a saturated solution. Thus in a gravimetric analysis the

precipitating reagent must be added to the solution of the sample until

KBD .of the precipitate is exceeded, and obviously the K V .

is most

readily exceeded if its value is very small; this is only another way of

saying that the analysis should be based upon the precipitation of a


"salt of the constituent being determined. It also

emphasizes the importance of basing the determination upon an irre-

versible chemical reaction, for it will be recalled that a reaction is

complete if an insoluble product is formed. Nothing, of course, is

absolutely insoluble, but the less the solubility (the smaller the KU9.)

the better the chances for a good quantitative separation of the desired


It should be pointed out that the relative solubilities of salts are not to be

judged directly from their K* values unless they are of the same valence types.

For example, silver chloride and silver chromate at room temperature are soluble

to the extent of about 0.0015 g. and 0.026 g. per liter respectively. Therefore the

Kav of silver chloride is

- 015>-

1.1 X 10-

and that of silver chromate is

Ag2CrO 4 , A'HP =

Thus, although silver chloride has the lesser solubility, it has the greater solubility

product. The reason obviously is that the solubilities of both salts is a fraction of a

mole per liter and the cubing of the fraction in the one case gives a smaller value

for Ks p. than does the squaring of the fraction in the other case. (In addition,

the larger value of the molecular weight of silver chromate contributes some-

what to its smaller solubility product constant.)


Common Ion Effect. The product of the ion concentrations in a

saturated solution is limited by the solubility product. Thus if to a

saturated solution of silver chloride an excess of silver ion is added the

solubility product is momentarily excx^ be pre-

cipitated until^the^roJuct of thejon^concgntrationsi^is sufficiently dimin-

ished tojroestablishJfche./f.^value. The addition of any compound having

arflon in commor^ witheither^

ion of the Ttifficultty sblutle~solute wilt

decrease dissolubility of the lattcfrthatjs tosay,jvvill

cause a more nearly

complete precipitation of the latter. This is the well-known common ion

effect. Because of this effect it is a rule wKeif precipitating a salt to~a33~a

slight excess of the precipitating reagent, since this will always give an

excess of a common ion. The amount by which the solubility is decreased

Page 243: quimica inorganica cuantitativa


may be illustrated with a saturated silver chloride solution to which

excess silver nitrate is added. Since [Ag+][Cl~] = /ffl .p.,



If silver nitrate is added until the solution is 0.01 M with respect to

Ag+ ion, we have

-]~ - 10-

and since the solubility of silver chloride is identical with the concentra-

tion of the Cl~ JQiLJtsj5Qlubility has been reduced from 10~ 5 to 10~8 or a

thousandfold. This was to be expected since the silver ion^oric^ntration

was increased a thousandfold. The same result would be obtained if weadded potassium chloride to a saturated solution of silver chloride until

the solution was 901M_with respect to CI~Jon, for then


[Ag+] s ^-_T= 10-*

The above illustration brings us to another generalization regarding

solubility of a salt, namely, that the solubility is measured by the con-

centration of the ion which is not in excess. If there are 100 chloride ions

and 1000 silver ions in a solution we must regard the solution as con-

taining only lOOAgCl. Similarly a solution containing 100 silver ions and

1000 chloride ions also must be thought of as containing only lOOAgCl.

The same reasoning of course applies when the units for expressing the

quantities are moles per liter, i.e., the concentrations.

The common ion effect is somewhat different for a salt of a different

valence type, e.g., BaA. If a saturated solution of silver chromate is made

0.01 M with respect to Ag+ iori, the solubility of the salt is decreased

to a greater extent than if the solution were made 0.01 M with respect to

the CrO*" ion (see Fig. 32). In the first case

and9 V 10 12

[CrOr] - *_ 4- 2.0 X 10-8 moles per liter

but in the latter case

Page 244: quimica inorganica cuantitativa




FIG. 31. Solubility of silver chloride in the presence of excess common ion.

(From Kolthoff and Sandell: Quantitative Inorganic Analysis. By permission of

The Macmillan Company, publishers.)

and _ [Ag+] _

= 7.1 X 10-'

The common ion effect with various excess concentrations of each ion is

shown graphically in Figs. 31 and 32 for silver chloride and for silver

Page 245: quimica inorganica cuantitativa




-log [Cr04=]

FIG. 32. Solubility of silver chromate in the presence of excess common ion.

chromate. The data for the latter figure are given in Table 18; those for

Fig. 31 the student should calculate for himself. Note the symmetricalnature of the silver chloride curve and the asymmetry of the silver

chromate curve. 3 In both instances, of course, the cusp of the curve

occurs when neither the positive nor the negative ion is in excess; i.e.,

the solubility of a salt is at a maximum when there is no common ion


Solubility in Solutions Without Common Ion Effect. The Salt

Effect. One might be inclined to say, judging from the usual common ion

8Fig. 32 is really two graphs joined together. On the left we have the change of

solubility of silver chromate with varying excess concentrations of Ag+ ion; on the

right we have the change of solubility with excess concentrations of CrO4"

ion. Whenneither ion is in excess, the pAg+ and pOO*"" values must be 3.80 and 4.10; therefore

the abscissas of the two adjoining graphs must show 3.80 on tbe one side coincidingwith 4.10 on the otber.

Page 246: quimica inorganica cuantitativa


Table 18


effect, that in gravimetric analysis it would be a good idea to add a large

excess of precipitating reagent inasmuch as the common ion effect thereby

would become more pronounced. However, as may be seen in Figs. 31

and 32, a small excess of common ion is almost as effective as a large

excess. Furthermore, as a rule, a difficultly soluble salt dissolves to a

greater extent in a solution containing a moderate or a high concentra-

tion of an electrolyte than it does in water alone, whether or not a

common ion effect is at play. In almost all of our applications of the

Law of Mass Action it is possible to employ the concentrations of the

ions involved instead of the active masses. This usually introduces no

serious error. Occasionally, though, it is necessary to avoid thia approxi-

mation if we are to understand certain unusual effects. It has alreadybeen pointed out that the active mass is represented by the activity and

not the concentration. The activity, a, and the concentration, c, are

related thus:

(6) a =fc

where / is the activity coefficient mentioned in Chapter 7, p. 105. As a

given solution becomes more dilute, / approaches unity, so that at

infinite dilution a and c become equal. The value of a for an ion in

moderately concentrated solutions depends upon the valence and uponwhat other ions are present in the solution, for interionic attraction will

vary as these conditions change. If we employ activities instead of con-

centrations for the difficultly soluble salt, BA, we have

(7) oB+.aA.- = SB

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where aB+ and aA~ represent the activities of the two ions and S repre-

sents the thermodynamic solubility product. Combining equations (6)

and (7)

(8) [B+][A-]/B+./A- = SBA


(9) pj+HA-]-^In a saturated solution of a salt only very slightly soluble in water, the

dilution is great and the values of /B f and /A- are almost unity; thus

the thermodynamic solubility product becomes sensibly equal to the

classical solubility product. However, in the presence of other electro-

lytes /B ^ and /A- decrease in value so that, in order that SBA of equation

(8) may remain constant, the values of [B+] and [A~] must increase;

that is to say, the solubility of the difficultly soluble salt, BA, increases

in a solution of an electrolyte beyond what it is in a pure water solution.

This is known as the salt effect.

Lewis and Randall4 have shown how the solubility product of a salt

in the presence of foreign ions may be obtained. Although the solubility

product of calcium sulfate, for example, is quite different when measured

in the presence of different concentrations of foreign salts (see Fig. 33),

there is a remarkable consistency revealed if the solubility product is

studied with variations in the ionic strengths, /* of the foreign salts.

The ionic strength is defined as half of the sum of the products of the

concentration arid the square of the valence of each ion present, or

(10) M = 0.5S(MZ 2)

where 2 is the summation symbol, M the molarity5 and Z the valence.

(For example, if the salt XY dissolves to the extent of 0.02 mole perliter in a 01 M solution of B 2A, the value of n is half the sum of (0.02)

(I)2 for X % (0.02) (I)

2 for Y~, (0.02) (I)2 for B+ and (0.01) (2)

2 for A~;hence 0.050.) If, now, the solubility product of calcium sulfate is

plotted .against ionic strength, as in Fig. 34, it is evident that in fairly

dilute solutions the solubility product is proportional to the ionic strengthno m^ter what the foreign salt be (contrast Fig. 33). Even at greaterionic Arength the variation of the solubility product is small; for example,

whenljLt = 0.1 it is seen fhat the solubility product of calcium sulfate in

the tmreo different solutions is 10~4 times 3.4, 3.4 and 3.5. The particular

valwl of the concept of ionic strength is that in dilute solutions the

4jLewis and Randall, Thermodynamics , The McGraw-Hill Book Co., New York,

1923^ pp. 373-383.

speaking, the molality.

Page 248: quimica inorganica cuantitativa


activity coefficient of a strong electrolyte is the same in all solutions of

the same ionic strength.

It should be noted that the salt effect manifests itself even when one

of the ions of the foreign salt is common with one of those coming from

the difficultly soluble salt. In such a case we have two opposing forces in

operation. When the excess of the common ion is not great, the result




0123456789Molanty of Foreign Salfx 100

FIG. 33. Change in the value of Kvv . for

calcium sulfate with change in molarity of for-

eign salts. (Courtesy, Rieman, Neuss andNaiman: Quantitative Analysis, New York,McGraw-Hill Book Co.)

will be the usual common ion effect and the slight excess causes a more

nearly complete precipitation of the difficultly soluble salt. But if the

excess ion is added to an extent much greater than 0.01 M, the salt effect

begins, as a rule, to predominate, and the solubility of the salt we wish

to precipitate is increased. Accordingly, one should avoid the addition

of a large excess of precipitating reagent in quantitative separations.

Complex Ions and Solubility. Sometimes it happens that the ions

of a precipitate will react with an excess of the precipitating reagent to

form a stable complex ion. Again, an ion from the precipitate may react

with another ion not common with either ion of the precipitated sub-

stance. In either case the formation of the complex ion must deplete,

Page 249: quimica inorganica cuantitativa


FIG. 34. Change in the value of 7v 8 .P .for calcium sulfate with change in the

ionic strength. (Courtesy, Rieman, Neuss and Naiman: Quantitative Analysis,

New York, McGraw-Hill Book Co.)

though not exhaust, one of the ions coming from the precipitate, and

this would tend to lower the ion product for the difficultly soluble com-

pound. Therefore, further solution of the compound must result to estab-

lish again its characteristic solubility product. Thus when a complex ion

is formed by the addition of excess precipitating reagent, instead of the

common ion reducing the solubility, an increased solution takes place.

The addition of excess ammonia to copper, nickel and silver salts, or of

sodium hydroxide to lead and aluminum salts, illustrates this type of

reaction. Such reactions often provide means of bringing about sepa-

rations in analytical chemistry; this has already been encountered in

qualitative analysis. The quantitative treatment of complex ion forma-

tion may be illustrated with the silver and cyanide ions. If some solid

silver argenticyanide is placed in pure water (or if a soluble cyanide and

a soluble silver salt are brought together in solution and silver argenti-

cyanide is precipitated), two equilibria are set up:


Ag[Ag(CN) 2] <=* Ag+ + Ag(CN) 2-, whence(solid)

[Ag+][Ag(CN)r] = K,

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Ag(CN)r <= Ag+ + 2(CN)-, from which

[Ag*][CN-] _[Ag(CN)r]

- "2

Ki and K2 have been determined as 2.2 X 10~ 12 and 1.0 X 10~21 re-

spectively. What now will be the effect of adding excess cyanide ion to

the solution where both of these equilibria prevail?

First it is necessary to show that practically all of the silver ion in

the solution comes from the first of the above dissociations. Obviouslyfrom equation (11) the silver ion concentration, and that of the argenti-

cyanide ion as well, is \/2.2 X 10"12 = 1.5 X 10~ 6. The silver ion con-

centration coming from the second of the two dissociations may be

calculated, since we know that if we let the value of [Ag+

]from this

source be X, then [CN~] = 2X, and [Ag(CN) 2~], we have already seen,

is 1.5 X 10~ 6. Therefore

(XX2X)!1.5 X 10- 6

" 1<0 X LU

from whichX or [Ag+] = 7.2 X 10" 10. Thus practically all of the silver ion

comes from the first equilibrium; the total value of [Ag*-]~ 1.5 X 10~ 6


Suppose that enough potassium cyanide is added to the solution to

reduce the silver ion concentration (by shifting the equilibrium, Ag(CN)^"+ Ag+ + 2(CN)~, to the left) from 1.5 X 10~ 6 to 1.0 X 10~ 6

. This elim-

inates a concentration of silver ion amounting to 0.5 X 10~~ 6 and causes

an equal increase of argenticyanide ion concentration. Momentarily,then, the ion product, equation (11), will be

[1.0 X 10- 6][(1.5 X 10- 6

) + (0.5 X 10- 6)]= 2.0 X 10~ 12

which is less than the solubility product for these ions namely,2.2 X 10~ 12~so that the precipitate must proceed to dissolve until the

K^. value is reestablished. Thus, because of complex ion formation, wehave the reverse of the usual common ion elFect. (What would be the effect

of adding excess silver nitrate to a precipitate of silver argenticyanide?)

Hydrogen Ion Concentration and Solubility of Salts. Difficultlysoluble salts of strong acids are neither more nor less soluble in strongacid solutions than they are in salt solutions of comparable ionic strength.But a difficultly soluble salt of a weak acid will dissolve if treated with a

strong acid. Consider calcium oxalate. The small amount which dissolvesin water dissociates completely:

CaC 2 4 * Ca+(solid)

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If a strong acid is added to a precipitate of calcium oxalate the hydrogenions will combine with oxalate ions, since oxalic acid is a weak acid and

because the ioriization of the oxalic acid which is formed is repressed bythe excess hydrogen ion from the strong acid. The result is that the value

of [C 2O 4~] is reduced to such a point that the ion product of [Ca+^tCaOr1


falls below KRV .for calcium oxalate; therefore the solid salt must dissolve

in its attempt to restore Ks . to its constant value. The newly formed

oxalate ion now combines with hydrogen ion, provided enough of the

strong acid was added, thus allowing for a continuation of the dissolving

process until, finally, all of the calcium oxalate may be dissolved. Ob-

viously, for solution to take place it is necessary that the value of the

anion concentration, after adding the strong acid, be less than that which

the salt itself would establish in the absence of acid. The higher the value

of /C P .for the salt and the smaller the value of Ka for the weak acid

which is formed, the easier may this be accomplished. Quantitative pro-cedures often take advantage of the solvent effect of acids, varying as it

must with the two factors just mentioned. The precipitation of silver

chloride, in the gravimetric determination of chloride or of silver, is

carried out in a slightly acidic solution. This prevents the simultaneous

precipitation of other anions, since silver salts, arsenates, carbonates,

chronmtes, oxalates, etc., are soluble in slightly acidified solutions. The

quantitative precipitation of salts which dissolve in acidic solutions must,of course, be carried out in a basic solution. Indeed, it may be said that

in almost every instance where the principle of solubility product is in-

volved the hydrogen ion concentration is a matter of utmost importance.Fractional Precipitation, Separation of Ions Which Are Pre-

cipitated by the Same Reagent. Usually the precipitation of a certain

ion in the form of one of its slightly soluble compounds takes place in a

solution containing several ions other than the one which it is desired to

determine. In other words, the purpose is to precipitate a single sub-

stance in a solution containing, as a rule, several ions of like charge; thus

the term, fractional precipitation. In only a minority of cases is it true,

when the precipitating reagent is added to a solution containing manyions of like sign, that the solubility product principle rigorously governsthe precipitation which occurs. There are a few, however, and as an

illustration we may consider the precipitation of silver chloride from a

solution containing both chloride ions and chromate ions. Here, both

anions behave independently; that is, neither is precipitated upon the

addition of silver ion until the product of the concentration of the anion

times that of the silver ion raised to the proper power exceeds the K^v .

value of the compound. This fact is the basis of the Mohr method for the

volumetric determination of chlorides which is described elsewhere in this

text (see p. 283),

Page 252: quimica inorganica cuantitativa


The solubility products of silver chloride and silver chromate are


[Ag+][Cl-] = 1 X 10-i


= 2 X 10- 12

If silver nitrate solution is added to a solution containing both chloride

ion and chromate ion, each of the above conditions must be satisfied at

the same time. If we square the equation for silver chloride and divide

by that for silver chromate, the silver ion cancels out and we obtain


2 X 10*

This is to say that when both silver chloride and silver chromate are in

equilibrium with the supernatant liquid the concentration of the chloride

ion when squared will be one two hundred millionth of the concentration

of the chromate ion. Obviously the concentration of chloride ion which

may remain in solution with chromate ion when the two are at equilibriumwith silver ion may be only very small; practically speaking, zero.

If, therefore, silver ion is added, say to equivalent quantities of chlo-

ride and chromate ions in the same solutionrsllVer chloride will precipitate

alone until the chloride ion is so reduced in concentration that the condi-

tions of equation (13) are realized. After this both silver chloride andsilver chromate will precipitate, and in this same ratio, since the state-

ment embodied in equation (13) may be maintained in the supernatant

liquid only if, once both begin to precipitate, they do so in this ratio.

Evidently in order that one substance be practically completely pre-

cipitated before the other begins to form it is necessary that the solubility

products of the two substances be considerably different in magnitude.Another example is a mixture of iodide and chloride which when dissolved

may be separated by the addition of silver ion, since Ka .v. for silver iodide

is 10~ 16 while that for silver chloride, as already mentioned, is 10" 10.

The student should calculate the relative amounts of chloride and iodide

in solution after both precipitates are in equilibrium with the solution.

Fractional Precipitation Regulated by Controlled pH. A dis-

cussion of buffer action is given in Chapter 7, p. 97. The means described

there for obtaining and maintaining a definite hydrogen ion concentrationmust be kept in mind in considering the following aspects of fractional


We recall from qualitative analysis the importance of carefully con-

trolling the pH of the soljution in the separation of the copper-arsenic

group of metals as sulfides from the metallic ions of subsequent groups.A saturated solution of hydrogen sulfide at room temperature is approxi-

Page 253: quimica inorganica cuantitativa


mately 0.1 M. If the hydrogen ion concentration is adjusted to about

0.3 M we have

[H+p[S-] _ (0.3)2[S-] 2

[H,S]~~07l *


(14) [S-]= 1.2 X 10- 22

Now the solubility product of copper sulfide is about 10~45 and that of

zinc sulfide is about 10~" 23. Thus, if precipitation of copper sulfide is to be

avoided when one of its salts is treated with a solution saturated with

hydrogen sulfide and 0.3 M with respect to hydrogen ion, the concentra-

tion of copper ion must be equal to or less than 10~23, i.e.,

whereas the similar calculation for zinc ion yields

in-23= 10-t

This means that with the hydrogen ion concentration so controlled (and,

therefore, also the sulfide ion concentration) only about 6.4 X 10~* 22 g.

per liter of copper, but 6.5 g. per liter of zinc, could remain in solution.

In other words, it would be difficult to prevent practically all of the copperfrom being precipitated, but any amount of zinc up to 0.1 mole, or 6.5 g.,

per liter would remain dissolved after treating with the hydrogen sulfide.

A different value for hydrogen ion concentration would establish a dif-

ferent sulfide ion concentration, for the latter obviously varies inversely

as the square of the former; accordingly, carelessness in regulating the

acidity in sulfide separations may easily result in the precipitation of

sulfides of iron, aluminum, etc., if these ions are present.

Likewise a strict control of hydrogen ion concentration is necessary

if ferric hydroxide is to be precipitated alone in the presence of magnesiumions. The solubility products for ferric hydroxide and magnesium hydrox-ide are 1.1 X 10~36 and 1.2 X 10-". If to a solution containing both

ferric ions and magnesium ions a properly buffered solution of ammonium

hydroxide is added, ferric hydroxide may be precipitated without throw-

ing out any magnesium hydroxide. Suppose we have a solution which is

0.05 M with respect to both Fe+++ and Mg++ and to it is added sufficient

ammonium hydroxide and ammonium chloride to make the solution 0.1 Min regard to each. The question is: What will precipitate, if anything?

The answer is simple once we know the value of [OH-]. This may be

Page 254: quimica inorganica cuantitativa


calculated from the dissociation constant of the base, namely, 1.8 X 10~~ 6,

and the concentration of the salt.6


X = [OH-]then

0.1 + X = [NH 4+1


0.1 - X = tNH 4OH]

Substituting the above in the mass law expression,

[NH our= 1>8 x 10

~6' we obtain

X 10- 6 or(o.i

~ X)~

[OH~] = X = 1.8 X 10~ 5

Therefore the ion product for ferric ion and hydroxyl ion in the solubility

product expression is

(5 X 10- 2)(1.8 X 10~ 6

)3 = 2.9 X 10- 16

which is greater than KKV. for ferric hydroxide, and therefore it precipi-

tates. At the same time the ion product for magnesium ion and hydroxylion in the solubility product expression is

(5 X 10--)(1.8 X 10- 5)2 = 1.6 X 10- 11

which is almost equal to the /C P . for Mg(OH) 2 . The fact that the ion

product is very slightly higher than the /CM) docs not mean that precipita-tion of magnesium hydroxide will certainly occur, since, due to the

tendency to form supersaturated solutions, it is necessary as a rule that

the ion product exceed the K* v. to a considerable extent. However,theoretically it would be necessary to use somewhat more ammoniumchloride than was stated in the above problem, thus further repressingthe ionization of the ammonium hydroxide, to be certain that no mag-nesium hydroxide would precipitate. The student should calculate

exactly what is the smallest possible concentration of the salt which

6 The [OH~], of course, may be computed from the buffer equation of Chapter 7p. 98. That is,

- (-j

(1.8 X ]

[OH-] 1.8 X 10-'

Page 255: quimica inorganica cuantitativa


would make the precipitation of magnesium hydroxide impossible.

(Answer: 0.12 M.)


(See also p. 246 )

1. Explain under what circumstances the relative solubilities of salts in grams

per liter may or may not be judged directly from their Ksp . values.

2. The common ion effect upon the solubility of a salt may be shown graphically

by plotting the solubility against the negative logarithm of the excess ion as in

Figs. 31 and 32. What types of salts will yield symmetrical curves?

3. Why does /, the activity coefficient, approach unity as the dilution becomes

very great P

4. Explain how the solubility of a precipitate is rendered greater (a) if one of its

ions reacts to form a complex ion, or (b) through the salt effect (see p. 227).

Page 256: quimica inorganica cuantitativa

Chapter iU


AN GRAVIMETRIC analyses when a substance is precipitated, washed,

dried and weighed the percentage of the essential constituent present

in the original sample may be calculated if^_and_onlyjft the precipitate

consists entirely of a compound of definite and known composition. In

accordance with the solubility product principle the analyst avoids

exceeding the /fs.p.of all compounds except that which he wishes to

precipitate. However, it is unfortunately a fact that very often during a

precipitation not only the desired substance is thrown out of solution,

but, in addition, quantities, usually small but significant, of other sub""

stances the solubility products of which have not been exceeded. This

phenomenon is known as coprecipitation, and the contamination of

precipitates resulting from coprecipitation is one of the most commoncauses of inaccurate analyses. Some compounds, such as silver chloride,

separate in a fairly pure form, but many others, barium sulfate^Jbr

example, do not do so under ordinary conditions. It is by no means

always a fact that the contaminating substance is one which itself is a

difficultly soluble substance. It is true that when copper sulfide is precipi-

tated from a solution also containing zinc ions a small amount of zinc

sulfide is carried down, and that when ferric hydroxide is precipitated

with ammonia from a solution also containing cupric ion some copper

hydrpxide is found to contaminate the precipitate. But highly soluble

substances also are coprecipitated in many instances. For example,

barium sulfate, given the chance, will drag down with it ferric sulfate or

barium nitrate, to mention only two, both of which are quite soluble.

The study of coprecipitation would be comparatively simple if there were

a consistent pattern of behavior always exhibited when a precipitate

shows an unexpected impurity. The fact is, however, that while a few

signs of conformity of behavior have been observed, to a large extent

each is a case to itself, so that it is at present difficult to make predictions

or set forth rules which would enable us, in advance of actual experimen-

tation, to say whether or not contamination of a certain precipitate is

likely to occur under a given set of conditions. Nevertheless, from the con-

siderable amount of data wvhich has been accumulated on the subject, we

often are able to say what appear to be causative factors, and to confine

the contamination within reasonable bounds, if no prevent it. When this


Page 257: quimica inorganica cuantitativa


is not possible an attempt may be made to arrange conditions so that

unavoidable positive or negative errors of coprecipitation are compensated

by intentional and nullifying negative or positive errors of about the

same magnitude. The precipitation of barium sulfate is an example of

such a compensation.1

j^The Colloidal State. Examples have already been encountered in

qualitative analysis when precipitates were obtained in such a finely

divided state that the particles passed through the filter. Often the oxi-

dation of hydrogen sulfide causes a precipitate of sulfur comprised of

such fine particles that they will not settle in a reasonable length of time

and cannot be filtered out of the solution. Such_particles are known_as a

temporary suspension; they are recognizable as a separate phase. ^The

diameter of such particles is about 10~B cm. Each consists of many mole-

cules, or atoms, since the diameter of a single molecule is around 10~ 7 or

10~8 cm. Under the effect of gravity the particles of a suspension eventu-

ally would settle. But if the particles happen to assume a size somewhatsmaller than 10~5 cm. a condition of affairs results which is quite different

from that prevailing in a suspension. No separate phase can be seen

(there is no cloudiness in the solution) ; however, a path of light throughthe liquid becomes visible when viewed transversely (Tyndall effect).

Furthermore, a separation of these colloidal particles from the solution

can be accomplished through dialysis, that is, by the use of a semi-

permeable membrane which will allow the passage of the molecules of a

true solution but will retain the colloidal particles. Thus a colloidal dis-

persion may be defined as a system in which the particles of the dispersed

phase are intermediate in size between those of a suspension and a true

solution, the latter being characterized by a subdivision of the solute

down to molecular or ionic magnitudes.It is not surprising that a colloidal dispersion should be quite different

in properties from both true solutions and suspensions. A true solution is

homogeneous, if not philosophically, at least practically speaking; a col-

loidal dispersion is heterogeneous, for there are two phases: the dispersed

phase and the dispersing medium. A suspension, given time enough, will

settle but from the solution, thus revealing its heterogeneity. A solution

exhibits coliigative properties which depend upon the number of parti-

cles: osmotic pressure, increase of boiling point, decrease of freezing

point and lowering of vapor pressure. Colloidal dispersions do not exhibit

these properties.

In analytical chemistry we are usually concerned only with colloidal

dispersions in which the dispersion medium is water. Such dispersionsare called hydrosols. They may be roughly grouped into two classes,

hydrophobic colloids, or those in which the particles do not show a

1 Fales and Thompson, Ind. Eng. Chem., Anal. Ed., 11, 206 (1939).

Page 258: quimica inorganica cuantitativa


strong affinity for the dispersion medium, water; and hydrophilic colloids,

or those whose particles do exhibit a strong affinity for water. Upon

coagulation the former retain relatively little water, while the latter

coagulate to yield gelatinous solids which carry a large amount of water;

in the coagulated form they are known as gels.

For the most part the hydrosols met in analytical chemistry are

electrically charged, either positively or negatively, due to the fact that

they adsorb either cations or anions from the solution in which the

particles are dispersed. Arsenious sulfide, for example, may be prepared

as a colloidal dispersion; the particles assume a negative charge by ad-

sorbing sulfide ions on their surfaces. Ferric hydroxide, as a colloidal

dispersion, takes on a positive charge. When colloidal dispersions are

electrolyzed they lose their electrical charge and become coagulated.

They also are coagulated by the addition of an electrolyte which will

furnish ions to serve as centers of attraction around which the colloidal

particles may cluster and thus flocculate. It is the reverse of this process

that is, the removal of the ions which serve to flocculate the colloidal

particles which accounts for the annoying experience of leakage through

the filter when a coagulated precipitate is washed free of foreign ions.

The precipitate, in part, simply reverts to a colloidal dispersion. Such

peplizalion can be prevented by washing, not with pure water, but with a

solution of an electrolyte which later, usually during the ignition of the

precipitate, may be removed.

Both hydrophobic and hydrophilic colloidal dispersions are met in

quantitative work. They are troublesome not only because they cannot

be filtered but also because, due to the large aggregate surface area which

such small particles must possess, they tend to attach relatively large

amounts of foreign ions to these surfaces. Even though coagulated they

retain much of the contaminating substances. It therefore is necessary

to control conditions so that precipitation is effected without any con-

comitant formation of a colloidal state. When the solubility product of a

salt in a true solution is exceeded a supersaturated solution results. Pre-

cipitation follows supersaturation and the crystallization of the substance

occurs about centers of nucleation within the solution. The development

of these centers depends chiefly upon the temperature and the presence

of foreign bodies. If the rate of nucleation is low while the rate of crystal

growth is high, the result will be the precipitation of fewer and larger

particles. A high rate of nucleation and a low rate of crystal growth pro-

mote many small particles and may even bring about a colloidal dis-

persion. This of course is undesirable since the presence of a great number

of very small particles, whether they actually attain colloidal dimensions

or not, means a large 'surface area which, in turn, leads to adsorption of


Page 259: quimica inorganica cuantitativa


J]</> U.0 O3 <0</> -o



Concentration of adsorbed substance in solution

FIG. 35. Coprecipitation as a function of the concentration of the contaminatingsalt.

Adsorption. Adsorption may be defined as the increase in concentra-

tion of a substance, molecule or ion, at the surface of a solid in contact

with a solution of those molecules or ions. If an aqueous, solution of

hydrogen sulfide is shaken with particles of activated charcoal, and the

solution is then filtered, it will be found that the odor is hardly noticeable.

The molecules of hydrogen sulfide have been attracted to the surface of

the charcoal particles and thus removed from the solution; they havebeen adsorbed. Considerable evidence exists which indicates that adsorp-tion frequently is the reason back of coprecipitation. Let us examinesome of this evidence.

Adsorption equilibria in general are found to follow the law formulated

by Freundlich,2

X - kCv*

where X is the weight of substance adsorbed per unit weight of adsorb-

ing solid, C is the concentration of the adsorbed substance in the solu-

tion, k and n are constants, the latter usually having a value from 1 to 5.

The graph for the above equation yields different curves depending uponthe values of the constants but, in general, takes a parabolic form as

seen in Fig. 35.

The Freundlich equation indicates that if the coprecipitated sub-

stance is adsorbed its quantity should be proportional to its concen-

tration in the supernatant liquid. That is, as the concentration of the

objectionable ion becomes greater, the amount coprecipitated will become

8 Freundlich, Z. physik. Chem., 57, 385 (1907),

Page 260: quimica inorganica cuantitativa


greater. There are many experimental confirmations of this. Blasdale3

precipitated barium sulfate from a series of equimolar solutions of sul-

furic acid containing differing amounts of potassium nitrate. With

0.2000 g. of potassium nitrate present the barium sulfate was found to

be contaminated with 0.0062 g. of potassium nitrate; with 1.0000 g. of

potassium nitrate in the solution the precipitate carried down 0.0182 g.

of the nitrate; and with 5.0000 g. of potassium nitrate originally present

in the solution 0.0354 g. of it was coprecipitated. If these data are plotted

in the manner of Fig. 35 the points fall very well on a curve represented

by the equation, X = kC^\ that is to say, the data fit the Freundlich

equation. The same kind of evidence for adsorption of the coprecipi-

tated ion is furnished by Lehrman, Been and Manes4 in their studies of

the coprecipitation of barium ion by aluminum, chromium and ferric


It appears that coprecipitation usually takes place, at least with

crystalline precipitates, simultaneously with the precipitation. In other

experiments by Blasdale 1.0000 g. of potassium nitrate was added to the

suspension of barium sulfate after it was precipitated and only 0.0024 g.

of potassium nitrate was found to contaminate the precipitate. On the

other hand Kolthoff and Moskovitz 6 found that cupric ion is adsorbed

by gelatinous ferric hydroxide precipitate to the same extent regardless

of whether copper chloride is added before or after the precipitation of

the ferric hydroxide. Furthermore, these investigators found that the

extent of coprecipitation of cupric ion with ferric hydroxide was not

altered by digestion of the precipitate even up to 69 hours. This rather

strongly suggests that adsorption of the coprecipitated ion is at play,

since, unlike the situation when dealing with crystalline precipitates, the

digestion of gelatinous precipitates does not greatly change the surface

area, and adsorption, being a surface phenomenon, should therefore re-

main constant.

Another point of interest, and one which indicates the frequency with

which coprecipitation is due to adsorption, is the fact that generally

those ions which are the worst offenders in coprecipitating with precipi-

tates of large surface area are ions of high valence type. Other things

being equal, many hydroxide precipitates will drag down larger amountsof sulfate ion, for example, than nitrate ion. This conforms to the fairly

well established fact that crystalline substances generally adsorb greater

quantities of multivalent ions.

Much of the evidence mentioned above seems to indicate that ad-

8 Blasdale, The Fundamentals of Quantitative Analysis, D. Van Nostrand Co.,New York, 1928, p. 138.

4 Lehrman, Been and "Manes, J. Am. Chem. Soc., 62, 1014 (1940).6 Kolthoff and Moskovitz, J. Phys. Chem. t 41, 629 (1937).

Page 261: quimica inorganica cuantitativa


sorption is the principal cause of coprecipitation. However, there are

other examples of contamination of a precipitate in which adsorption

per se is not the primary factor.

Occlusion. Adsorption of an impurity during the growth of a crystal

is called occlusion. The contaminant may be adsorbed upon the surface

of the small crystals of the desired precipitate as they are being formed,

without entering into and becoming a part of the crystal lattice, and as

the crystals grow the contaminating ions are entrapped within the crystal.

It does sometimes happen that a foreign ion may actually replace an ion

of the same sign in the crystal lattice; this leads to the formation of a solid

solution. Thus if crystals of calcium oxalate monohydrate are agitated

with dilute solutions of potassium sulfate,6 some of the oxalate ions of

the crystal lattice are replaced by sulfate ions; shaken with a dilute solu-

tion of barium chloride, some of the calcium ions are replaced by barium

ions. The less soluble the compound formed between one of the ions in

the lattice of the precipitate and the foreign ion of opposite sign, the

greater the likelihood that the latter ion will enter the crystal lattice of

the precipitate by replacement. Schneider and Rieman7 have shown this

to be generally true in their study of the coprecipitation of certain anions

by barium sulfate. Of the six ions studied they found that the extent of

occlusion varied inversely with the solubility of the corresponding barium

salt with the exception of nitrite ion. Thus while the solubilities of the

barium salts of I~, Br~, Cl~, C1O 3~~ and N03

~ions decrease in the order

named, the extent of coprecipitation of these ions with barium sulfate was

exactly in the reverse order. Nitrite ion proved to be an exception,

possibly because it fits into the barium sulfate lattice better than the

other anions mentioned.

It is clear that in cases where occlusion is known to be the cause of

coprecipitation, digestion of the precipitate should decrease the extent

of occlusion. This is true because during digestion ions from the surface of

the crystals are continuously leaving the solid surfaces while ions from

the saturated solution are redepositing. Furthermore, as has been men-tioned already (p. 8), digestion brings about fewer and larger crystals.

While these changes are taking place foreign ions to some extent are

eliminated from the crystals and the contamination of the precipitate

therefore is decreased.

It frequently happens in cases where occlusion is known to take placethat the order in which two reacting solutions are brought together

governs the question of what the occluded ion will be. When a solution

containing the anion of the precipitate is added slowly to one containingthe cation of the precipitate, the occlusion of anions takes place to a much

Kolthoff and Sandell, J. Am. Chem. Soc., 55, 2170 (1933).7 Schneider and Rieman, J. Am. Chem. Soc., 59, 354 (1937).

Page 262: quimica inorganica cuantitativa


greater extent than does the occlusion of cations. Bringing the reagents

together in the reverse order causes the occlusion of excess cations. Whensulfate ions are added slowly to a solution of barium chloride the crystals

of barium sulfate are formed in an environment containing excess barium

ions. Thus a small crystal, immediately after being formed, starts to grow

by attaching barium ions which largely will find their counterparts in

sulfate ions as soon as more solution containing sulfate is added. If this

addition is slow, however, the counterpart to some barium ions of the

newly forming row or layer of the lattice will be made up of chloride ions

of which there is an abundance. Then as more sulfate ions are added and

the crystal of barium sulfate grows larger, some of the chloride ions find

themselves trapped and embedded beneath the layer of alternating

barium and sulfate ions in the now larger crystal. Thus when the growthof the crystal takes place while the cations (of the precipitate) are in

excess in the solution (that is, when the reagent slowly being added con-

tains the anion of the precipitate), anion occlusion predominates.

Schneider and Rieman7 demonstrated this type of selective occlusion

by barium sulfate by pouring a solution of sodium sulfate into a solution

of barium chloride and then reversing the process. The amount of chloride

ion occluded when the former method was carried out was found to be

six times greater than when the latter was employed.

Postprecipitation. As the name implies, this term denotes the con-

tamination of a precipitate by an impurity which comes out of solution

after the primary precipitation has been effected. Braithwaite and Fales 8

showed that when copper sulfide is precipitated from a solution contain-

ing zinc and cupric ions, and the precipitate is digested, the amount of

zinc sulfide contaminating the copper sulfide increases with the time of

digestion. Another example of postprecipitation is encountered whencalcium oxalate is precipitated from a solution containing magnesiumions. As usually carried out in the laboratory in the determination of

calcium, sufficient oxalate ion is added to exceed not only the solubility

product of calcium oxalate but probably that of magnesium oxalate as

well. 9Nevertheless, only a small amount of magnesium is found in the

precipitate. This is because the rate of precipitation of magnesium oxalate

is very slow. If, however, the precipitate of calcium oxalate is digested

for a day or so, much of the magnesium oxalate will have precipitated.

In general, if the amount of the impurity increases with the length of

time the precipitate remains in contact with the mother liquor, we ob-

viously are dealing with postprecipitation.

8 Fales and Kenny, Inorganic Quantitative Analysis, D. Appleton-Century Co.,New York, 1939, p. 241.

9 See Pedersen, Trans. Faraday Soc., 35, 276 (1939).

Page 263: quimica inorganica cuantitativa



It is often impossible to pi-event errors due to coprecipitation, but it

is usually possible to reduce such errors to reasonable limits. Before out-

lining the means of limiting the extent of coprecipitation it should be

pointed out that the error caused by such contamination is not neces-

sarily positive. If an ion of the impurity actually enters the crystal lattice

of the desired precipitate, and if that ion for example, sodium is lighter

than the one it replaces, the resulting error obviously will be negative.

If an ammonium salt contaminates a precipitate it will be volatilized

during the ignition of the precipitate. Therefore it may cause no error

at all if its coprecipitation was simple adsorption, for example; but if

the ammonium ion excluded an equivalent quantity of cation of the

precipitate then a negative error will result since the weight of the

former is so small.

Since the adsorption of impurities follows the Freundlich equation,

coprecipitation by adsorption is minimized by keeping the concentration

of the salt which is coprecipitated as low as possible. In the first place,

if the sample for analysis does not dissolve in water it should be broughtinto solution without introducing an ion known to cause trouble. For

example, if the choice of a solvent were between hydrochloric and nitric

acids and if the precipitate is known to adsorb or occlude nitrate ion it

would be unwise to use nitric acid to dissolve the sample. When, once in

solution, it is not practicable to remove the objectionable substance bysuch means as evaporation or precipitation, then the solution should be

made as dilute as permissible before the precipitating reagent is added.

If the precipitating reagent itself is coprecipitated, then it too should

be diluted before its addition, and it should be added slowly and with

stirring in order to prevent localized regions of high concentration duringthe precipitation. In some cases the contaminated precipitate is dissolved

and reprecipitated. The repeated precipitation results in far less copre-

cipitation of the impurity since the concentration of the contaminant in

the second solution from which reprecipitation takes place is much less

than in the original solution. Reprecipitation is feasible in the case of

hydrous oxides (hydroxides) which may be redissolved easily, but some

precipitates, barium sulfate for example, do not lend themselves to

"double precipitation" since they do not readily dissolve in any of the

usual solvents.

Since adsorption is a function of surface area, and since digestion

causes the formation of fewer and larger crystals, coprecipitation due to

surface adsorption or to occlusion usually may be reduced by heatingthe precipitate in contact with the mother liquor. Furthermore, adsorp-

Page 264: quimica inorganica cuantitativa


lion in general decreases with elevated temperature; therefore digestion

at higher temperatures is advantageous.

It has already been mentioned that the order in which the reacting

substances are brought together is sometimes important. The procedure

for carrying out any routine analysis usually has been worked out so

that a minimum of coprecipitation is encountered. In investigating a new

method involving a precipitation one should be on the alert for copre-

cipitation errors. In any new study it must be remembered that just

because an ion is or is not coprecipitated under one set of conditions is

no reason that it will or will not contaminate even the same primary

precipitate under another set of conditions. Once the facts regarding the

existence and the extent of coprecipitation have been established, means

possibly may be found to correct the situation. If not, and if the error

exceeds allowable limits, a new procedure must be developed.

Questions and Problems

1. Explain the circumstances under which a salt precipitate will be dissolved bya strong acid.

2. Explain why the addition of an electrolyte to the wash water may prevent the

peptization of a precipitate. What two particular properties must the electro-

lyte possess in order that its use will introduce no error in the analysis?

3. What effect will a change of temperature have on the solubility product of

the majority of salts?

4. Why, as a rule, is the filtration of a precipitate an easier matter if the pre-

cipitating reagent is added slowly and with stirring than if added all at once?

5. Explain the advantage of digesting a precipitate prior to filtration. Under

what condition would a long period of digestion be a disadvantage?

6. Show that a slightly soluble salt of the A 2B 3 type has a solubility product

equal to 4X5/9 where X is the concentration of the B~ ion.

7. The compound AB 2 is twice as soluble at a temperature of 30 as it is at 20.

How much greater will its solubility product be at the higher temperature?Answer: 8 fold greater.

8. If 0.0025 g. of the compound A 2B 3 , having a molecular weight of 250, satu-

rates a liter of solution, what is its solubility product constant?

Answer: 1.1 X 10-.

9. The slightly soluble compound Afi 2 has a molecular weight of 200 and is

soluble to the extent of 5.0 X 10~ 4 moles per liter, (a) Calculate the solu-

bility product, (b) How many milligrams of the compound remain dissolved

per 100 ml. of solution if an excess concentration of A++ ion equal to 0.01 Mis added? (c) How many milligrams of AB 2 will be lost if the precipitate is

washed with 100 ml. of pure water?

Answer: (a) 5.0 X 10~ 10; (b) 2.2 mg.; (c) 10 mg.

10. (a) The solubility product of the slightly soluble hydroxide, Me(OH) 2, is

1.28 X 10~ 12. If a solution of sodium hydroxide is slowly added to an acidified

solution which is 0,02 M with respect to Me++ion, at what pH will the

Page 265: quimica inorganica cuantitativa


Me(OH) 2 begin to precipitate? (b) If 1.07 g. of ammonium chloride is addedto 50 ml. of a solution which is 0.4 M with respect to NH 4OH, and this is

poured into 50 ml. of solution which is 0.0079 M with respect to Me++ ion,

will Me(OH)2 precipitate or not? (Assume a total volume of 100 ml. after

pouring the two solutions together.)

Answer: (a) 8.9; (b) no precipitate, saturated solution.

11. At what pH will a 0.030 M solution of ferric sulfate begin to precipitate

Fe(OH) 3 ? #,.p. for Fe(OH) 3 = 1.1 X 10~36.

Answer: 2.5.

12. Calculate the pH of a buffer solution which will allow the precipitation of all

the cobalt except 0.1 mg. when 250 ml. of a 0.05 M solution of cobalt chloride

is saturated with hydrogen sulfide at room temperature. K*.v . for CoS = 1.9

X 10- 27.

Answer: 0.70.

13. Calculate the solubility, in g. per 100 ml., of silver chromate in a 0.005 Msolution of silver nitrate and in a 0.005 M solution of sodium chromate. /C,.p

for Ag 2Cr0 4= 2.0 X lO' 12


Answer: 2.7 X 10~ 6; 3.3 X 10" 4


Page 266: quimica inorganica cuantitativa

Chapter 15



J.HE CHLORIDE is dissolved, the solution acidified with dilute nitric

acid and the anion is precipitated with silver nitrate. The precipitate of

silver chloride,

Cl- + Ag+ -+ AgCl j

is coagulated by digesting and then is filtered, washed, dried and weighed

to a constant weight (see p. 15). The precipitate can be filtered through a

filter paper and ignited to burn the paper to a "weightless" ash; this

results in the reduction of a large part of the silver chloride to silver bythe carbon of the paper and necessitates the reconversion of the reduced

silver to silver chloride by treatment with aqua regia. This procedure is

not recommended since the use of the Gooch crucible, or any type of

filtering crucible, eliminates the extra step necessitated by the reduction

occurring during ignition. With the Gooch crucible the precipitate of

silver chloride may be filtered, washed, dried and weighed immediately.

The method given for the determination of chloride obviously maybe employed for the determination of silver; in addition, it may be ap-

plied to the determination of any ion the silver salt of which is insoluble

in solutions slightly acidified with nitric acid. These include the halides,

bromide and iodide (the silver salts of which are even less soluble than

silver chloride) and cyanide and thiocyanate. Furthermore, hypochlorite,

chlorite, chlorate, hypobromite, bromate, hypoiodite, iodate and per-

iodate ions may be reduced with sulfurous acid to the corresponding

halide, after which they too may be precipitated as the silver halide

and the analysis carried out according to the procedure given below.

Obviously, if any one of these ions is to be determined by this method,

all others (that is, cyanide or thiocyanate, or any other halide) must be


Procedure. Prepare Gooch crucibles or porous filtering crucibles as

described on p. 10, dry them at 105 and weigh to a constant weight

(see p. 15). Accurately weigh out samples of between .0.2 and 0.4 g. of

the dry, finely divided chloride and transfer quantitatively to 250 ml.

beakers. Dissolve in 150 ml. of chloride-free water and add 8 or 10 drops

of concentrated nitric acid. Calculate the volume of 5 per cent silver


Page 267: quimica inorganica cuantitativa


nitrate solution necessary to precipitate the chloride in the sample,

assuming it to he pure sodium chloride. To the cold solution, and with

constant stirring, add this amount of silver nitrate plus a 5 per cent excess

or a 3 ml. excess (see p. 224), whichever is the greater. Cover with a

watch glass and heat until almost boiling; then transfer to a hot plate

or a steam bat/i and digest until, after stirring, the precipitate quicklysettles. Then test for completeness of precipitation by adding a drop or

two of silver nitrate solution; add more silver nitrate if a fresh precipi-

tate appears in the test.

Place the weighed Gooch crucible in the suction flask and connect to

the vacuum pump (see Fig. 1). Then decant the supernatant liquid into

the Gooch crucible and transfer the precipitate to the crucible as de-

scribed on p. 10. When the transfer has been completed, wash the pre-

cipitate in the crucible, still using suction, with a dilute solution of nitric

acid prepared by adding 1 ml. of 6 TV nitric acid to 500 ml. of water. In

filtering and in washing never Till the crucible more than three-quartersfull. After six or eight such washings collect a wash portion which has

just passed through the crucible in a small test tube; test for silver ion

by adding 1 or 2 drops of dilute hydrochloric acid. The faintest cloudiness

indicates the need for further washing. Before discarding the filtrate

make a final test for completeness of precipitation by adding 1 ml. of

silver nitrate solution. No further precipitate should be obtained, but if

one does appear an excess of silver nitrate must be added and the newlyformed precipitate digested and filtered through the same Gooch crucible.

When the precipitate is clean, as shown by a negative test for silver

ion, place the crucibles with tongs in clean, dry 250 ml. beakers and dryin the oven at 105 for an hour or more. Cool in the desiccator and weighto the fourth decimal place. Return to the oven at 105 for 30 minutes,then cool arid weigh again. Repeat the 30-mimite drying period and the

weighing as many times as is necessary to obtain a "constant" weight;that is, until two successive weighings are within 0.3 ing. of one another.

From the weight of the precipitate of silver chloride calculate the per-

centage of chlorine in the sample. Duplicate results should show a devi-

ation from the mean no greater than 3 parts per 1000.

Notes. In computing the amount of silver nitrate solution necessary to precipi-tate all of the chloride in the sample, it is safest to base the calculation on the

assumption that the unknown is sodium chloride. Since sodium is one of the light-est of metals, it follows that the quantity of silver nitrate necessary to precipitatethe chloride in a given weight of sodium chloride will be more than sufficient to

precipitate that in the same weight of the chloride of a heavier metal. In this waythe risk of later discovering that too little precipitating reagent had been added

originally is practically eliminated. However, if it is thought that the unknowncontains only a small percentage of chloride it is better to make no calculation ofthe amount of silver nitrate to be added, but rather to add the reagent until no

Page 268: quimica inorganica cuantitativa


further precipitation is visible, thus avoiding the salt effect (see pp. 227-230)

which might introduce a negative error.

Test the distilled water to see whether or not it is chloride-free by adding to a

few milliliters of the water in a test tube a crystal of silver nitrate. A cloudiness

indicates the presence of chloride ion. In that case consult the instructor.

The solution of the chloride is acidified with nitric acid to prevent the precipi-

tation of other anions which might be present, the silver salts of which are insolu-

ble in neutral solutions but which are soluble in acid solutions (e.g., carbonates

and phosphates). The solution of the chloride must not be heated, once it has been

acidified, until after an excess of silver nitrate has been added; otherwise a loss of

chlorine will result because of either or both of the reactions: Cl~ + ll+ > HC1,

and 6H+ + 6C1~ + 2HNO3- 3C1 2 + 2NO + 4H 2O.

The decomposition of silver chloride is catalyzed by light. In direct sunlight

the action is rapid. Therefore the precipitate should be shielded from bright light.

The superficial bluish color which the precipitate assumes is due to the metallic

silver from this decomposition, but in diffused light the effect on the weight of the

precipitate is not significant. Note that if the suspension of the precipitate is ex-

posed to bright light before being filtered, the error will be negative if the solution

surrounding the precipitate contains excess chloride ion (as in a determination of

silver), but will be positive when the solution surrounding the precipitate con-

tains excess silver ion (as in a determination of chlorine). In the former case the

decomposition due to light results simply in a loss of chlorine, but in the latter

situation the chlorine from the decomposition of silver chloride reacts with excess

silver ion, giving additional silver chloride. The net gain then will be a silver atom

for every molecule of silver chloride which was decomposed and then re-formed.

Since this experiment is often the first precipitation-gravimetric determination

which the student performs, the importance of washing the precipitate until free

from contamination is frequently not fully appreciated. The slightest cloudiness

indicates the need for further washing. Though other ions are present in the liquid

from which the precipitate is filtered, it may be assumed safely that when the

wash solution shows no test for silver ion the precipitate is also clean in respect to

other ions.

Washing is carried out with water slightly acidic with nitric acid in order to

prevent peptization, as explained on p. 11. The small amount of nitric acid

remaining with the precipitate after the washing is completed is volatilized dur-

ing the drying which follows. So also would hydrochloric acid, which could be

used in dilute solution for the washing if silver and not chlorine were being de-

termined; in this case it has the advantage over a nitric acid wash solution of

decreasing the solubility of the precipitate due to the common ion effect.

Tt is better never to fill the Gooch crucible entirely full of liquid. This will

insure a clean portion of the crucible where it may be grasped with the tongs

without loss of precipitate.

The initial period for drying the crucible and precipitate in the oven may be

shortened if, after washing the precipitate, two rinses of alcohol are passed

through the crucible followed by two of ether.

Calculation of Percentage of Chlorine. The precipitate which is weighed

consists of silver chloride, AgCL Since there is one atom of chlorine with an atomic

weight of 35.46 in each molecule of silver chloride, molecular weight L43.3, it

follows that chlorine constitutes 35.46/143.3 of the weight of the precipitate. This

fraction of the weight of the precipitate divided by the weight of the original

sample and multiplied by 100 will give the percentage of chlorine.

Page 269: quimica inorganica cuantitativa


Suppose, for example, a chloride sample weighing 0.2023 g. yields a precipitateof silver chloride weighing 0.4951 g. The percentage of chlorine in the sample then


(35.46)(0.4951)(100) _(143.3) (6:2023)"



The principle involved in the gravimetric determination of sulfur is

quite simple, being the precipitation of the sulfur in the form of barium

sulfate, filtering the precipitate and finally weighing the barium sulfate.

However, few gravimetric determinations are so fraught with possible

errors. Barium sulfate tends to coprecipitate (see pp. 238-246) manycontaminants, and the error thus introduced may be either positive or

negative, depending upon the nature of the impurity coming out with

the precipitate. Anions like chloride, chlorate and nitrate, if present whenbarium sulfate is precipitated, are coprecipitated as the correspondingbarium salts in varying amounts depending upon the conditions (con-

centration, order of bringing reagents together, etc.), and cause highresults. Many cations also may be coprecipitated; the alkali metals and

calcium and ferric iron are brought down with the barium sulfate in the

form of their sulfates. Coprecipitated cations almost always cause low

results if sulfate is being determined because the molecular weight of

the contaminating sulfate is lower than that of barium sulfate. If ammo-nium ion is present when barium sulfate is precipitated the negativeerror is especially pronounced because the ammonium sulfate which is

coprecipitated will be completely volatilized upon ignition (the final step

before weighing). During ignition the composition of many coprecipitated

substances changes; for example, ferric sulfate is converted to ferric oxide,

and this always causes low results in sulfate determinations. The fact

that fairly accurate results may be obtained comes from careful attention

to the details of a procedure which has been worked out with two ends

in view namely, to reduce, as much as possible, coprecipitation of

foreign substances, and to effect a compensation of positive and negativeerrors which cannot be avoided entirely.

Because barium sulfate, practically speaking, is insoluble in all sol-

vents, the process of reprecipitation as a means of purifying the precipi-

tate cannot be employed. When it has been established that certain ions

known to coprecipitate with barium sulfate are present in the sample for

analysis, steps should be taken when possible to remove such ions before

the barium sulfate is precipitated. For example, nitrate ion may be re-

moved by acidifying the solution of the sample with hydrochloric acid

and evaporating. Errors due to the presence of ferric ion may be reduced

in three ways. The iron can be removed by precipitation as ferric hy-

Page 270: quimica inorganica cuantitativa


droxide and the sulfate then determined in the absence of ferric iron.

The ferric iron may be reduced to ferrous iron and the barium sulfate

precipitated in the presence of the latter. This reduces the error since

ferrous iron does not coprecipitate with barium sulfate so badly as does

ferric iron. Again, the ferric iron can be converted into a complex ion by

tartaric acid and thus reduce the extent of coprecipitation.

A great deal has been published regarding the errors commonly en-

countered in the gravimetric determination of sulfur (or barium) and the

various methods of combating or minimizing these errors. A compre-

hensive survey of the matter may be found elsewhere. 1 The procedure

given below is suitable for the analysis of a soluble sulfate in a solution

not containing significant concentrations of the worst offenders amongthe coprecipitating ions. Calcium and ferric iron, and nitrates and chlo-

rates must be absent.

Procedure. Accurately weigh out samples of about 0.5 g. and quanti-

tatively transfer to 600 ml. beakers. See that the beakers have no

scratches inside which would become impregnated with precipitate; also

make sure that stirring rods have smooth ends. Add 50 ml. of water and

if the sample does not completely dissolve add concentrated hydrochloric

acid dropwise and warm until the solution is clear. Add 2 ml. excess of

concentrated hydrochloric acid and dilute the solution to about 350 ml.

Calculate the amount of 1 per cent barium chloride solution necessary

to precipitate the sulfate in the sample, assuming it to be pure sodium

sulfate; to the hot solution of the sample add this amount plus a 5 per

cent excess, quickly and with rapid stirring. While still hot place over a

steam bath and after the precipitate has settled add 1 ml. of the barium

chloride solution to test for completeness of precipitation. If necessary,

add more barium chloride, stir, digest, allow to settle and test again.

When certain that an excess of the precipitating reagent has been added,

place on the steam bath or over a low flame for an hour or longer.

When the supernatant liquid is clear decant it through an ashless

filter paper. Test this filtrate with 1 ml. of barium chloride solution;

do not conclude that all of the sulfate has been precipitated unless no

cloudiness appears after standing 15 minutes; barium sulfate precipitates

only slowly from very dilute sulfate solutions. Place a new beaker under

the funnel (then if a leak occurs there*will be less to refilter) and transfer

the precipitate to the filter paper as directed on p. 10. The last bit of

precipitate may be removed from the beaker by using a policeman) or

the following method may be employed: With a piece of ashless filter

paper about the size of a postage stamp wipe off the film of precipitate

from inside the beaker and drop the paper into the funnel. Two or three

1 See Kolthoff and Sandeli, Textbook of Quantitative Inorganic Analysis, The Mac-millan Co,, New York, 1943, Chap, XX.

Page 271: quimica inorganica cuantitativa


pieces of paper should be sufficient to clean the barium sulfate entirely

from the beaker, and from the stirring rod as well.

Wash the precipitate about 10 times with hot water, each time filling

the filter cone almost full of water and allowing complete drainage before

adding the next wash portion (see p. 12). Collect a small amount of

washings in a test tube and add a few drops of silver nitrate solution.

If a precipitate forms, the washing must be continued until finally the

barium sulfate is free of chloride ion. It then may be assumed that the

precipitate is clean.

Place the funnel containing the filter paper and precipitate in the

oven for a few minutes; when partially dried remove the paper from

the funnel and carefully fold into a bundle, place in a previously weighed

porcelain or platinum crucible, char and ignite to a white ash. Thedetails of these several steps are described on p. 14. During the ignition

some of the barium sulfate might have been converted into barium oxide

or barium sulfidc. To remedy this add to the precipitate with a dropper

enough 1:10 sulfuric acid to dampen the entire mass (about 8 or 10

drops). Then with a very low flame located some 10 cm. underneath the

crucible cautiously warm until sulfuric acid furnes slowly are given off.

(Haste makes waste here; if spattering occurs because of too high a tem-

perature the entire determination will have to be repeated.) When all of

the excess acid has been expelled, elevate the flame until the bottom

half of the crucible glows at a faint red heat for 1 minutes. Then transfer

to the desiccator to cool. Weigh the crucible and contents and repeat the

10-mimite heating treatment, cool and weigh again. Repeat the heatingand weighing until a constant weight is obtained. From the weight of

the precipitate of barium sulfate calculate the percentage of sulfur, either

as such or in terms of sulfur trioxide, in the sample. While a precision of

1 part per 1000 can be expected with some experience, a deviation of

3 parts per 1000 may be considered excellent for the beginner. The pre-

cision may well exceed the accuracy. If the sample contains ions which

are particularly troublesome in coprecipitating, the error may amount to

15 or 20 parts per 1000.

Notes. The solution of the sample is acidified in order to prevent the simul-

taneous precipitation of other barium salts which are insoluble in neutral solu-

tions, e.g., carbonate or phosphate. The solution must not be made too acidic

since barium sulfate is somewhat soluble in dilute hydrochloric acid.

If the barium chloride is added slowly to the sulfate solution the occlusion of

alkali sulfate predominates and this, as has been explained already, leads to low

results. When the barium chloride solution is added rapidly some barium chloride

is occluded, but while this causes a slight increase in the weight of the precipitate,

the net result is quite close to the theoretical.

The crystals of freshly precipitated barium sulfate are extremely small ; diges-

tion causes the formation of larger crystals (see p. 8) and decreases the prob-

Page 272: quimica inorganica cuantitativa


ability of leakage during the filtration. After filtering off the precipitate examine

the filtrate carefully for small amounts of barium sulfate. Swirl the filtrate vigor-

ously with a stirring rod; if any precipitate is present it may be detected as a

"whirlwind" effect in the center of the rotating liquid.

During the ignition of the precipitate the decomposition of barium sulfate will

take place to some extent if the temperature is too high: BaS0 4 > BaO + S0 3 .

Furthermore, carbon from the filter paper may reduce some of the sulfate: BaS0 4

+ 2C -> BaS + 2C0 2 . Both BaO and BaS are reconverted to BaS04 by the

treatment with sulfuric acid:

BaO + H 2S0 4-* BaS0 4

BaS + H 2S0 4- BaSO 4 + II 2S


For the determination of sulfur in an insoluble sulfide e.g., FeS2

the sample must be rendered soluble either by treating with such re-

agents as aqua regia or bromine, or by fusion with such agents as sodium

carbonate combined with potassium nitrate, or sodium peroxide. In

either method the sulfide is oxidized to sulfate. In the former or wet

method the reaction may be represented as

FeS 2 + 15Br + 3HN0 3 + 8H2O - Fe(NO 3 ) 3 + 2H,SO 4 + ISHBr

In the fusion, or dry method, using sodium peroxide, the reaction is

2FeS 2 + 15Na 2 2 -> 4Na2SO4 + Fe2 3

In the wet method the oxidation takes place slowly and actually

may be incomplete; and proper precautions are necessary to prevent or

diminish coprecipitation resulting from the presence of nitrate and ferric

iron (see discussion under the preceding section). The dry attack has

some objectionable features too: large quantities of sodium or potassium

are introduced and great care must be taken to avoid sulfur being taken

up during the fusion from the gas of the burner. The latter may be

avoided, of course, by electrical heating. The procedure given below

utilizes the wet method and follows the principles of the Allen-Bishop


Procedure. Accurately weigh dry, finely divided samples of pyrite of

about 0.5 g. into 250 ml. beakers. Place the beakers containing the

samples in a hood and to each add 4 rnl. of bromine and 6 ml. of carbon

tetrachloride and cover the beakers with watch glasses. Keep at room

temperature for 15 minutes, stir occasionally and then add 10 ml. of con-

centrated nitric acid. Allow to stand again for 15 minutes with occasional

stirring. Next, place on a steam bath and heat below 100 until the action

has ceased and no more bromine fumes are evident. Remove the cover

8 Allen and Bishop, 8th Intern. Congr. Applied Chem., 1-2, 48 (1912); Moore, J. Ind.

Eng. Chem., 11, 46 (1919).

Page 273: quimica inorganica cuantitativa


from the beaker and continue heating on the steam bath until the solu-

tion has evaporated to dryness. Stir in 10 ml. of concentrated hydro-chloric acid and heat at 105 in the oven for an hour. This will dehydratethe silica. Remove from the oven and add 2 ml. of concentrated hydro-chloric acid, stir for a moment, and then add 50 ml. of water. Bring the

solution to the boiling point and let simmer for 5 minutes. Cool, and add

2 g. of aluminum powder in order to reduce the iron. After the solution

becomes colorless, cool, filter and wash the filter thoroughly. Receive

the filtrate and washings in a liter beaker and dilute to 700 ml. Add2 ml. of concentrated hydrochloric acid and 1 drop of stannous chloride

to prevent atmospheric oxidation of the ferrous iron and heat the solu-

tion to boiling. Precipitate the sulfate by adding 50 ml. of 5 per cent

barium chloride solution dropwise and allow the precipitate to settle for

an hour or longer. Then continue with the procedure as described for the

determination of sulfur in a soluble sulfate. Calculate the percentage of

sulfur in the ore. The deviation from the mean should be under 4 parts

per 1000.


Procedure. Accurately weigh samples of about 1 g. of powdered coal.

Mix with 3 g. of Eschka's compound, consisting of 2 parts of magnesiumoxide and 1 part of anhydrous sodium carbonate. Place the mixture in a

platinum crucible and cover with an additional 2 g. of Eschka's com-

pound. Protect the crucible from the flame of the burner by a shield andheat the crucible. Better still, heat the crucible in an electric furnace to

avoid completely contamination by sulfur-containing fuel gases. Appl>the heat gradually to expel volatile matter. Stir the mixture occasionallywith a platinum wire. After 30 minutes increase the heat to dull redness.

Remove the crucible and cool when no carbon remains and the color of

the mass is brown or yellow.

Place the crucible and contents in 100 ml. of water and heat to the

simmering point for 30 minutes. Remove the crucible and rinse it, catch-

ing the rinse water in the beaker containing the solution. Then filter

through a filter paper and wash the residue thoroughly, receiving the

filtrate and washings in a 400 ml. beaker. Add 5 ml. of saturated brominewater and 1 ml. of 6 N hydrochloric acid. Boil gently for 10 minutes.

(The bromine oxidizes any sulfites to sulfates.) When the solution is

colorless cool the solution, add 2 drops of methyl orange and then dilute

ammonium hydroxide until the solution is orange in color. Then add1 ml. of 1 N hydrochloric acid and proceed with the precipitation byadding dropwise 20 ml. of 5 per cent barium chloride to the hot solution

and continue as described for the determination of sulfur in soluble sul-

fates. Calculate the percentage of sulfur in the coal. For coals of about

Page 274: quimica inorganica cuantitativa


2 per cent sulfur content the absolute deviation for duplicate determi-

nations may amount to 0.05 per cent.



The gravimetric determination of phosphorus is accomplished b>

precipitating, from a solution alkaline with ammonium hydroxide, the

hexahydratc of magnesium ammonium phosphate, MgNH4PO4.6H 20.

Although the precipitate may be dried at room temperature after wash-

ing with alcohol and ether, and weighed, the majority of analysts prefer

to convert the precipitate by ignition into magnesium pyrophosphatcand weigh in that form. The equations for the precipitation and ignition


POr + Mg++ + NH 4+ + 6H 2O -> MgNH 4P0 4.6H 2O

2MgJNH 4PO 4.6H 2O -> Mg 2P 2O 7 + 2NH, + 13IT 2O

If the solution of the phosphate is acidic the value of [P0 4"J will be

relatively small since phosphoric acid ionizes in three steps and the third

ioni/ation constant is quite small. It follows that the concentration of the

phosphate ion will be affected greatly by the pll of the solution and that

the solubility of the magnesium ammonium phosphate will depend uponthe hydrogen ion concentration. The precipitation therefore is accom-

plished by adding magnesium ion to the acid solution of the sample and

then increasing the pH with ammonium hydroxide to the point whore

magnesium ammonium phosphate precipitates. However, the pH must

not be sufficiently high for magnesium hydroxide to precipitate. There-

fore, instead of simply adding a solution of a magnesium salt and ammo-nium hydroxide to the phosphate solution, "magnesia mixture" and

ammonium hydroxide are used. The former consists of magnesium and

ammonium chlorides and furnishes both of the cations needed to precipi-

tate the magnesium ammonium phosphate. Because the ammonium ion

decreases the hydroxyl ion concentration when the base is added, mag-nesium hydroxide does not precipitate.

The hydrates of magnesium hydrogen phosphate, MgHP04 , and the

normal salt, Mg3(PO4) 2 , are prone to separate out in significant amounts

along with the MgNH 4P0 4.6H 20, unless the ionic concentrations are

carefully controlled. If the precipitate is to be weighed as such without

ignition to the pyrophosphate, a procedure sometimes followed, it is evi-

dent that a pure precipitation must be brought about. This may be done

by dissolving the precipitate in acid and reprecipitating from the newsolution wherein the concentrations of the ions, including hydrogen ion,

are kept within narrower limits.

Page 275: quimica inorganica cuantitativa


Since nearly all metallic ions yield precipitates with ammoniacal mag-nesium chloride, it is imperative that before the magnesium ammonium

phosphate is precipitated all cations other than magnesium, ammoniumand the alkalies must be absent. In the analysis of phosphate rock, ferti-

lizers and the like it is necessary therefore to separate the phosphorusfrom interfering ions. This is brought about best by precipitating the

phosphorus as ammonium phosphomolybdate from a nitric acid solution.

P043 + 3NH4

+ + 12Mo04- + 24H+ -> (NHOaPO^l^MoOs + 12H 2O

The "yellow precipitate" does not always correspond to the formula

given in the above equation; therefore this precipitate is not weighedfor the estimation of phosphorus. However, since it contains all of the

phosphorus it is the means of separating the element from many inter-

fering ions such as invariably are present in phosphate ores. The yellow

precipitate can then be dissolved in ammonium hydroxide and mag-nesium ammonium phosphate then precipitated and ignited as already

described. When highest accuracy is not necessary the yellow precipitate

may be dissolved in a measured excess of standard sodium hydroxide

solution and the excess base determined by titration with a standard

acid solution. Fertilizers, steel and alloys are so analyzed for phosphorus.The reaction with sodium hydroxide is:

(NH4) 3P04.12Mo0 3 + 23NaOH -> llNa 2Mo0 4 + (NH 4) 2Mo0 4 +NaNII 4HPO4

or simply

(NH4) 3P0 4.12MoO 3 + 230H- -> 3NI1 4+ + 12MoO 4

~ + HPO 4= +

11H 2

In the analysis of a phosphate sample which docs not contain interfering

ions it is not necessary, of course, to precipitate the molybdatc; rather

the magnesium ammonium phosphate may be precipitated directly.

Procedure. Before beginning the procedure prepare the following

solutions which will be needed during the course of the analysis:

MAGNESIA MIXTURE. Dissolve 50 g. of magnesium chloride hexa-

hydrate and 100 g. of ammonium chloride in 500 ml. of water. Addammonium hydroxide until slightly alkaline and filter after 24 hours.

Acidify with hydrochloric acid and add 5 ml. of concentrated hydro-

chloric acid in excess. Dilute to 1 1.

AMMONIUM MOLYBDATE. Make up each of the following solutions:

(a) Dissolve 120 g. of molybdic acid (80 per cent Mo0 3) in a solution of

80 ml. of concentrated ammonium hydroxide in 400 ml. of water, (b) Add400 ml. of concentrated nitric acid to 600 ml. of water. Pour solutions

Page 276: quimica inorganica cuantitativa


(a) and (b) together, let stand overnight or longer, decant from any sedi-

ment and preserve the clear solution in a glass-stoppered bottle.

ALKALINE CITRATE SOLUTION. Dissolve 5 g. of citric acid in 140 ml.

of water and add 70 ml. of concentrated ammonium hydroxide.

Accurately weigh samples of about 0.25 g. of the dry, finely divided

material for analysis. Place in 150 ml. beakers, add 5 ml. of concentrated

hydrochloric acid and 2 ml. of concentrated nitric acid. Cover the beakers

with watch glasses and heat with a very low flame or on a hot plate

until only a few milliliters of viscous liquid remains. Then add 0.2 g. of

boric acid and 10 ml. of 1 : 10 nitric acid. Heat until the solution boils and

filter through a 7 cm. filter paper into a 250 ml. beaker. Wash the residue

with five successive portions of 10 ml. of water each, catching the wash-

ings in the beaker containing the filtrate.

To the filtrate slowly add concentrated ammonium hydroxide until a

slight precipitate is produced. Then add concentrated nitric acid drop-

wise and with stirring (use a thermometer as a stirring rod) until the

precipitate just redissolves; then heat the solution to 60 and add 75 ml.

of warm ammonium molybdate reagent. Digest the precipitate which

forms for 30 minutes at 60, stirring frequently. Put aside at room tem-

perature and as soon as the precipitate has settled sufficiently test for

complete precipitation by adding 1 ml. of the molybdate reagent to the

clear, supernatant liquid. If additional precipitate forms, add more mo-

lybdate solution, digest and test again. Allow the precipitate to stand in

contact with the solution for 4 hours, after which filter and wash the

precipitate by decantation with three 10 ml. portions of 5 per cent ammo-nium nitrate solution. Place the beaker which still contains nearly all of

the precipitate underneath the funnel and dissolve the precipitate by

pouring a total of 20 ml. of the alkaline citrate solution through the filter.

Then wash the filter with three portions of warm 1:20 ammonium hy-

droxide, and three portions of 1:10 hydrochloric acid. The total volume

at this point must not exceed 150 ml. Add 2 drops of methyl red indicator

and then concentrated hydrochloric acid until the color changes from

yellow to pink; then add 1 ml. of the acid in excess. (Add the acid slowly

and with stirring: if the yellow precipitate re-forms add more ammonium

hydroxide and neutrali/e with acid again.)

Cool the slightly acid solution in ice and add 10 ml. of magnesiamixture and then concentrated ammonium hydroxide solution slowlyand with stirring until the solution is alkaline. Now add 5 ml. in excess.

The white precipitate which appears is magnesium ammonium phosphate

hexahydrate. Allow it to settle several hours or overnight.Filter through a filter paper and wash three or four times by decanta-

tion with cold 1:20 ammonium hydroxide, retaining most of the precipi-

tate in the beaker. Then place this beaker under the funnel and wash

Page 277: quimica inorganica cuantitativa


the filter with 50 ml. of warm 1 : 10 hydrochloric acid. Follow with five

washings of the filter with small portions of 1 : 20 hydrochloric acid. Fromthe cold combined filtrate and washings reprecipitate the magnesiumammonium phosphate hexahydrate by adding first 2 ml. of magnesiamixture and then freshly filtered ammonium hydroxide, the latter 5 ml.

in excess. Allow the precipitate to settle several hours or overnight, then

filter and wash thoroughly with 1 : 20 ammonium hydroxide.

Partly dry the filter and precipitate and then transfer to a weighed

platinum or porcelain crucible. Ignite over a moderate flame until no

carbon is visible. Place the crucible in a inullle furnace at 1100 or heat

it with a Meker burner for 1 hour. Cool in a desiccator and weigh.

Repeat the heating for half-hour periods and the weighing until a con-

stant weight is obtained. Calculate the percentage of phosphorus in

terms of P2C>5. In experienced hands the deviation from the mean of

duplicate results will be under 5 parts per 1000; for the student runningthe analysis for the first time a deviation of 10 parts per 1000 is acceptable.

Notes. Phosphate rock usually contains fluorine; boric acid is added to the

acid solution of the phosphate to convert fluorides to fluoboric acid, HBF 4 , in

which form fluorine does not interfere.

Ammonium phosphomolybdate precipitates tend to go into the colloidal state,

but the tendency is less if the precipitation is effected at 60. Do not heat much

higher than 60 or the ammonium molybdate will yield molybdic anhydride,

MoOs, which then largely comes out with the desired precipitate.

The yellow precipitate is washed with a solution of ammonium nitrate to

prevent pcptization of the ammonium phosphomolybdate.The ammonium citrate solution converts iron and tin into soluble complex

compounds, thus preventing the precipitation of these metals as phosphates.The double precipitation of magnesium ammonium phosphate is necessary

because of the presence of molybdate in the initial precipitate.


The following method for determining phosphorus avoids the precipi-

tation of ammonium phosphomolybdate. Calcium, the chief metallic con-

stituent in phosphate rock, is kept in solution by the presence of citrate

ion, and the phosphorus is precipitated as magnesium ammonium phos-

phate. Aluminum, calcium, fluorine, iron, manganese, titanium, silica,

zinc and organic matter, in the quantities found in phosphate rock, do

not interfere. A large excess of magnesia mixture must be added because

the citrate ion converts a good part of the magnesium ion into a complex.

Pyrophosphate also may be determined by this method.

Procedure. Accurately weigh out samples of the dry finely groundrock weighing about 0.3 to 0.5 g. into 250 ml. Krlenmeyer flasks. To each

add 10 ml. of concentrated hydrochloric acid and 2 ml. of concentrated

8 See Hoffman and Lundell, J. Research Natl. Bar. Standards, 19, 59 (1937).

Page 278: quimica inorganica cuantitativa


nitric acid. Cover the flasks with watch glasses and simmer for 30 minutes.

Rinse off watch glasses, add 25 g. of ammonium citrate, 5 ml. of hydro-

chloric acid and then 75 ml. of the solution of magnesia mixture (p. 257).

Add concentrated ammonium hydroxide carefully until the solution is

neutral to litmus and then 2 ml. in excess. Dilute to 200 ml. with water,

add a few glass beads, stopper and shake frequently during a 30-minute

period. Then allow to stand overnight. Filter and wash the filter paper

and flask once with 1:20 ammonium hydroxide. Discard the filtrate.

Dissolve any precipitate still in the flask with 50 ml. of 1 :4 hydrochloric

acid; then pour the acid through the filter paper to dissolve the precipi-

tate on the paper. Wash the flask and the paper with small portions of

more 1:4 hydrochloric acid totaling about 50 ml. Add 0.3 g. of citric

acid and 1 ml. of magnesia mixture to the solution. Add, while stirring,

concentrated ammonium hydroxide until the solution is alkaline to litmus

and then 5 ml. in excess. Stir frequently during a 30-minute period and

then allow to stand for 4 hours or overnight.

Filter the precipitate of magnesium ammonium phosphate through

an ashless filter paper and wash thoroughly with 1:20 ammonium hy-

droxide. Partially dry the precipitate and paper, transfer to a weighed

platinum or porcelain crucible, char the paper at a low temperature arid

ignite to a constant weight with a Meker burner or in a muffle furnace.

Calculate the percentage of phosphorus present in terms of P 2O 5 . The

precision should be as good as that obtained by the procedure of the

ammonium phosphomolybdate method (p. 259).


After bringing the material into solution, iron may be determined

by precipitating it as ferric hydroxide, or rather as hydrous ferric oxide,

Fe 2O3.xH 2O. The precipitate is filtered on paper and ignited to ferric

oxide, Fe2O 3 ,in which form it is weighed.

Hydrous ferric oxide is extremely insoluble; its incipient formation is

of a colloidal nature but the presence of foreign ions causes it to coagulate

into a gelatinous mass so that it can be filtered. The filtration will be

slow, however, unless the coagulation is maintained. Washing by decan-

tation, prior to filtration, is advisable, and a dilute solution of a volatile

ammonium salt should be used in the washing. This will prevent the

dispersion of the coagulated gel.

Hydrous oxides such as that of iron, being of a colloidal nature,

possess a vast surface area so that they show a strong tendency toward

the adsorption of foreign ions. Precipitated from an alkaline solution,

hydrated ferric oxide adsorbs hydroxyl ions which, in turn, adsorb a

secondary layer of cations. Contamination therefore must be prevented

by double precipitation; this may be done since the precipitate ferric

Page 279: quimica inorganica cuantitativa


oxide may be redissolved in acids. The second precipitation will yield a

much less contaminated product (see Coprecipitation, p. 245).

Ignition of the hydrated ferric oxide requires particular care if accu-

rate results are to be obtained. The paper must be charred slowly or the

ferric oxide will be partly reduced to magnetic iron oxide, Fe.3O 4 . A goodcirculation of air during the ignition will aid in preventing this. In addi-

tion, a sufficiently high temperature, close to 1000, must be maintained

to drive off the last trace of water which is rather tenaciously held.

Metals like aluminum, trivalent chromium and titanium must be

absent since they also are precipitated by ammonium hydroxide. Anions

such as arscnate, phosphate, vanadate and silicate cannot be tolerated

for they form insoluble iron salts in weakly alkaline solutions. Further-

more, there are some other substances which interfere with the determi-

nation because they prevent the precipitation of iron by forming complexions with the ferric ions for example, organic hydroxy compounds such

as tartaric and citric acids. These, however, may be destroyed byoxidation.

Procedure. The sample may be fairly pure iron itself, a soluble iron

compound such as ferrous ammonium sulfate or an ore of iron. In the

last case dissolve the sample in a manner described in the procedure for

determining iron volumetrically (p. 191), but do not add any stannous

chloride. If iron wire or filings comprise the sample it may be dissolved

in dilute hydrochloric acid.

Accurately weigh samples of about 1 g. of ferrous ammonium sulfate,

or of iron ore, or about 0.2 g. of iron wire or filings. Bring into solution,

dilute to 50 ml., heat to boiling and add dropwise 1 ml. of concentrated

nitric acid. Continue to boil gently for 5 minutes. The initial brown color

is due to the formation of FeSO4.NO, but this decomposes with the

boiling and a yellow color appears.

Dilute the solution to 200 ml., heat almost to the boiling point andadd freshly prepared 1:1 ammonium hydroxide (filter before usingif any silica is present in the reagent bottle), while stirring constantly.An excess of ammonium hydroxide is assured it* the odor of ammonia

persists just above the beaker when the solution is boiling gently. Con-

tinue to boil for a minute or two after adding the hydroxide, then removethe flame and allow the hydrated ferric oxide precipitate to settle some-

what. Test the supernatant liquid by adding a few drops of ammonium

hydroxide; note whether or not any additional precipitate is formed. If

so add more ammonium hydroxide.

Decant the supernatant liquid through an ashless fast-filtering paper;retain the bulk of the precipitate in the beaker and wash by decantation

four times with 50 ml. of hot water. Discard the washings and replace

the beaker under the funnel with the one containing the bulk of the pre-

Page 280: quimica inorganica cuantitativa


cipitate. Pour 50 ml. of 1 TV hydrochloric acid into the filter paper so

that it continuously wets the entire paper and then wash the filter with

hot water until it is perfectly white. Remove the beaker containing the

precipitate, the acid and washings, and heat. If necessary to bring all of

the precipitate into solution add more 1 N hydrochloric acid. Add a few

drops of nitric acid to insure all the iron being in the ferric state. Repre-

cipitate the hydrated ferric oxide by adding ammonium hydroxide as was

done the first time. After the precipitate settles filter through the same

filter paper as was used before, again washing by decantation but with

hot ammonium nitrate solution, 4 g. to 500 ml. of water. Finally bring

all of the precipitate into the filter paper (see p. 11) and continue to

wash the precipitate and paper with ammonium nitrate solution until a

test portion gives no precipitate with a few drops of silver nitrate solution.

When the precipitate has been washed clean1

, partly dry it by placing

the funnel in the oven at 105 for (only) 10 minutes; then transfer it to a

previously weighed porcelain crucible as directed on p. 15. Char the

paper slowly and then ignite with the full heat of the burner, taking

care to exclude the gases of the flame from the interior of the crucible.

After 20 minutes no carbon should remain. Cool in the desiccator and

weigh; then reheat for 15-minute periods until a constant weight is ob-

tained. Calculate the percentage of iron in the sample in terms of Fe 2O 3 .

The average deviation from the mean result should be 5 parts per 1000

or less.

Notes. The 1 ml. of nitric acid is added in order to oxidize any ferrous iron to

the ferric state. Should the precipitate come out with a greenish instead of a

brick-red color, ferrous iron is indicated. In that case dissolve the precipitate by

adding a little hydrochloric acid, and add another 1 ml. portion of nitric acid

and boil gently for several minutes. Then reprecipitate by adding ammonium

hydroxide.The precipitate of hydrated ferric oxide must be washed immediately after

filtering. If allowed to dry, the impurities become calked in the precipitate and

are very difficult to wash out.

Hydrous metallic oxides are generally peptized by water. The first washing of

the precipitate may be done with water (and the water may be hot since the pre-

cipitate has such a low solubility), for there are sufficient foreign ions present to

prevent peptization. After the second precipitation, however, it is advisable to

wash with a solution of ammonium nitrate.

The precipitate may be regarded as free from all contaminating ions when it

has been washed free of chloride ion. Thus the chloride ion test, which is simple

to make, may be used to indicate the elimination of impurities.


In analyzing a substance in regard to one particular constituent, as

has been done in preceding experiments, any error which might be made

detracts from the accuracy, but only in regard to the one constituent.

Page 281: quimica inorganica cuantitativa


In the analysis of a substance such as limestone, and later, of brass, in

which the composition in regard to a number of constituents is sought,

it is evident that an error made at one point does not merely vitiate the

result for the constituent being separated at the moment, but, in addi-

tion, might contribute to the error of all subsequent determinations. It

follows that all of the proficiency which the student by now has come to

command should be brought to bear upon each step as, one by one, the

several constituents are separated and determined.

Limestone consists principally of calcium carbonate. It is the raw

material used in the manufacture of lime and its analysis is widely per-

formed in routine control work. In addition to calcium carbonate, lime-

stones usually contain varying amounts of magnesium and smaller

proportions of silica, iron, aluminum and manganese, and often phos-

phorus, sulfur, titanium and the alkalies. A scheme for the complete

analysis of limestone may be found in Hillebrand and Lundell, Applied

Inorganic Analysis, John Wiley & Sons, New York, 1929; in the "proxi-mate" 4

analysis which follows, those constituents or components are

determined which ordinarily are run in commercial laboratories. These

include (a) loss on ignition (water, carbon dioxide and any organic

matter), (b) silica, (c) R2O3 (the combined oxides of aluminum, iron,

titanium, manganese and phosphorus), (d) calcium oxide, (e) magnesiumoxide and (f) carbon dioxide.


The sample must be collected and prepared for analysis in accord

with the general rules already mentioned (see pp. 5-6). Accurately weigh

samples of not more than 1 g. into platinum crucibles which previously

have been ignited over the Meker burner or in a muffle furnace at 1000

to 1100 and weighed to a constant weight. Place in the drying oven at

105 to 110 for 2 hours, cool in the desiccator and weigh. Repeat the

drying for 1-hour intervals and the intervening weighings until constant

weight is obtained. Report the loss of weight as due to hygroscopic water.

After the determination of moisture the same samples may be used

for the determination of loss on ignition. When the limestone is heated

to around 1000 or 1100 the principal change taking place is the loss of

carbon dioxide and "essential hydrogen" (hydrogen which in combined

form is an integral part of the molecule) in the form of water. In addition,

4 The determination of the elementary composition of a complex substance (suchas limestone) is called the ultimate analysis of the substance, even though the con-

stituent elements might be reported in terms of their oxides, for example. The proxi-mate analysis involves the determination of the major constituents as radicals, or

compounds, or even as mixtures e.g., as moisture, ash, mixed oxides, etc. The term

proximate docs not imply any departure from the usual practices for attaining high

precision and accuracy.

Page 282: quimica inorganica cuantitativa


any organic matter which may be present is decomposed, and oxidations

such as of sulfides to sulfates and ferrous iron to ferric iron will occur.

The last two oxidations will cause a gain in weight; however, the loss in

weight of a dry sample usually will give a measure of the carbon dioxide

content with an accuracy of around 1 per cent.

Procedure. Use the residue from the moisture determination or weigh

out new samples into platinum crucibles which previously have been

heated to 1000 or 1100, cooled and weighed to a constant weight.

Cover the crucibles and heat over a Tirrill burner at a dull red heat for

5 minutes and then increase the heat to the full capacity of the burner

for 5 minutes. Then replace the Tirrill burner with a Mcker burner or

blast lamp, or place in a muflle furnace at 1000 to 1100. After 30 minutes

cool in the desiccator and weigh; then heat again for 10-minute intervals,

cool and weigh until a constant weight is obtained for each crucible. If

the residue from the moisture determination was used, include the per-

centage of moisture in the result for loss on ignition. Keep the residue

for subsequent determinations. Duplicate determinations should check

within 0.2 per cent.

Notes. Porcelain crucibles may be used for the ignition but platinum crucibles

are preferable. Porcelain will change weight at the high ignition temperature

(platinum inappreciably). It is a good idea to rcweigh the crucible after the

material has been removed and the crucible cleaned, and to use this weight as the

weight of the empty crucible. Porcelain crucibles, being rather thick, have the

further disadvantage of being difficult to raise to the high temperature required if

a Meker burner is used instead of the muffle furnace.

The crucible should be heated at a lower temperature at first in order to

avoid a too rapid initial evolution of carbon dioxide. This might blow out fine

particles of the limestone and cause erroneous results both here and in subsequent


After ignition the material is quite hygroscopic. Therefore the crucible should

be kept covered and the weighing made as rapidly as possible (see p. 63).


The silica in the original sample of limestone may be present as such

and in the form of insoluble silicates. The latter are rendered soluble by

igniting with salts of sodium, potassium, calcium, etc., usually with

sodium carbonate (p. 7). In the case of limestone the amount of calcium

oxide obtained during the ignition of the sample is usually sufficient to

convert the insoluble silicates into acid-soluble silicates. Since, further-

more, the quantity of silica in limestones generally is low, the addition

of a flux is not necessary as a rule; but when the silica content is ex-

ceptionally high the proportion of calcium oxide may be insufficient to

convert the silica into acid-soluble form. In such cases a fusion witji

sodium carbonate is necessary. The resulting sodium silicate (Si02 +

Page 283: quimica inorganica cuantitativa


Na2C0 3 Na 2Si03 + C02) may then be treated with hydrochloric acid,

forming a precipitate of silicic acid. Dehydration of the silicic acid is

necessary before it is filtered. This may be accomplished either by

evaporating the mixture of silicic acid and silica to dryness and heating

at 105 as in the procedure below, or by fuming with perchloric acid.

When it is treated with water or dilute acid the gel tends somewhat to

pass into a colloidal state. A double evaporation and dehydration at

105, however, will diminish the dispersed phase to negligible propor-

tions; that which escapes filtration is largely compensated for by small

amounts of metallic oxides resulting from coprecipitation.

Procedure. Transfer the residue from the determination of loss on

ignition to 150 ml. beakers. Add 1 or 2 ml. of 1:1 hydrochloric acid to

the crucible to loosen any pHV|fring p^rtinlos and add to the solution in

the beaker, being careful that none is lost in the transfer. Repeat the

washing of the crucible with small portions of dilute acid until the quanti-tative transfer of all of the residue to the beaker is assured. Keep the

total volume as small as possible. Add 5 ml. of concentrated acid to the

beaker. Warm and stir the mixture until the disintegration is completeand then evaporate over a steam bath until dry. This may be hastened

by occasionally breaking up the pasty mass with a stirring rod as it

approaches dryness. When a dry, mealy powder is obtained place in the

drying oven at 105 for an hour. Then remove from the oven and moisten

the residue with 5 ml. of concentrated hydrochloric acid, and 5 minutes

later add 25 ml. of water and heat on the steam bath to dissolve the

acid-soluble substances present. Filter the residue of silica through a

7 cm. ashless filter paper and wash four times with hot 1:100 hydro-chloric acid and then with hot water until free from chlorides, catching

the filtrate in 400 ml. beakers. Unless the silica present is less than 5 or

10 mg., evaporate the filtrate to dryness, dehydrate and repeat the treat-

ment with hydrochloric acid, the filtering and washing exactly as before.

This double dehydration will regain the silica which went into solution

during the first extraction. Reserve the filtrates and washings for suc-

ceeding determinations. Ignite the combined silica residues from the two

dehydrations in platinum crucibles already weighed to constant weight,

using Meker burners or a muffle furnace at 1000. After obtaining con-

stant weights calculate the percentage of crude silica in the sample. The

duplicate results should not differ by more than 0.2 per cent.

Notes. A small amount of silica escapes filtration even with double dehydra-tion. This loss is just about counterbalanced as a rule by coprecipitated iron* andaluminum oxides. In the most exact analysis two corrections are necessary: (a)

the silica which remained in solution after the above procedure was carried out

must be recovered from the precipitate of "combined oxides" (see final note of

next section) ; and (b) the amount of pure silica in the above crude silica must be

Page 284: quimica inorganica cuantitativa


estimated. The latter may be done by treating the crude, moistened silica with

3 drops of 1 : 1 sulfuric and 5 ml. of hydrofluoric acids; the mixture is then evapo-

rated and ignited in the platinum crucible to convert the metallic sulfates to

oxides, and reweighed. The loss in weight obviously represents pure silica, since

all SiO 2 was expelled in the form of SiF 4 during the evaporation and ignition which

followed the hydrofluoric acid action' (2H 2F 2 + Si0 2 -> SiF4 | + 2H 20). The

residue from the hydrofluoric acid treatment should be fused with a very small

amount of sodium carbonate and the cake dissolved in hydrochloric acid and

added to the main filtrate.


Ammonium hydroxide, added to the filtrate from the silica determi-

nation, precipitates the "combined oxides,^" consisting of hydrous oxides

of iron, aluminum, titanium and manganese, and the phosphates of these

metals. Any silica not previously removed will be precipitated here. The

manganese is completely precipitated only if first oxidized; this is accom-

plished witli bromine or ammonium persulfate and at tfie same time iron

is oxidized to the ferric state. If the manganese were not precipitated

here as hydrous manganese dioxide it would come out later with calcium

or magnesium. Since aluminum is amphoteric, a large excess of ammo-

nium hydroxide cannot be tolerated. A double precipitation is necessary

for combined oxides because some calcium and magnesium are always

coprecipitated. After redissolving the hydrous oxides in hydrochloric

acid, they are reprecipitated from solutions obviously containing only

little of the calcium and magnesium originally present. The second pre-

cipitation therefore yields the mixed oxides practically free from these


Procedure. The filtrates from the silica determination should have a

volume of 150 to 200 ml. Add to each about 1 g. of ammonium chloride

so that sufficient ammonium io^ willjbe present to pfe

tation of magnesium when ammonium hydroxide is ^

sorutiOTrtt/Eoilh^g-and add 2 riiirbT saturated bromine Jwte?and simmer

for 3 minutes. To the hot solution add 2 drops of methyl red indicator

and then carbonate-free concentrated ammonium hydroxide from a newly

opened bottle (filter if necessary) until the solution is slightly alkaline.

Boil the solution gently for 2 or 3 minutes and filter immediately through

a small, ashless filter paper into 400 ml. beakers. Wash the precipitate

twice with hot water, then place the filtrate to one side for future use.

Place the original beaker under the funnel and pour over the filter paper

25 ml. of 3 AT hot hydrochloric acid to dissolve the precipitate. Wash the

paper five times with hot water. To the solution in the beaker, which

should have a volume of around 75 ml., make a second precipitation of

the hydrous oxides, using bromine water and ammonium hydroxide as

before. Collect the precipitate on the same paper as was used after the

Page 285: quimica inorganica cuantitativa


first precipitation and receive the filtrate in the beaker containing the

first filtrate. Wash with 1 per cent ammonium nitrate solution until a

small wash portion gives a negative test for dhjorides! Do not test for

cloudiness until after washing five times, and then use only 1 ml. for the

test. (Begin the evaporation of the combined filtrates in preparation for

the determination of calcium.) Place the filter paper containing the

hydrous oxides in a weighed porcelain crucible, char the paper and ignite

over a Mekqr burner and weigh to a constant weight. Calculate tlw

percentage of "R 203" in the sample. The results of duplicate analysis

should agree within 0.2 per cent.

Notes. Bromine water is preferable to ammonium persulfate for oxidizingthe manganese and iron in the solution since the use of the latter contributes sul-

fate ions. This would result in the coprecipitation of sulfate with the calcium

oxalate when calcium is determined. The manganese is oxidized by bromine to

MnC>2 which upon heating becomes Mn 3Oi.<r>

More than a slight excess of ammonium hydroxide in precipitating the hy-drous oxides is objectionable for three reasons: it would cause some of the hydrousaluminum oxide to dissolve; it would react with bromine (2JNH 4

f + 3Br 2 > N 2

+ 6Br~ + 8H f) ; and it would permit the precipitation of some calcium carbon-

ate, since some carbon dioxide from the air is bound to enter the solution.

The ammonium hydroxide used should be carbqnate-freeFor an ultimate analysis the elements present in the mixed oxides may be

individually determined by dissolving the residue in hydrochloric arid, or fusingwith potassium bisulfate, and separating in a manner similar to the procedure in

qualitative analysis. If desired, the small amount of silica which escaped separa-tion up to this point may be determined by dissolving the potassium bisulfate

fusion of l\i>O:j in 1:10 sulfuric acid, evaporating to dehydrate the silica, taking upwith dilute acid to dissolve the metallic sulfates and filtering oft the silica. Ignite

and weigh this residue and add to the corrected weight (after hydrofluoric acid

treatment) of silica previously obtained.


At this stage in the analysis of limestone the solution (the combined

filtrates from the R20a determination) contains calcium, magnesium, po-

tassium and sodium cations, a great deal of the last if a sodium carbonate

fusion was necessary in the beginning. Calcium is determined by precipi-

tating calcium oxalate from an almost rTeutral" solution. \n easily filtrable

precipitate is obtained by acidifying the solution, adding, ammmiiiimoxalate and then gradualty~Tricreasing the pH by adding ammonium

hydroxide. Unfortunately both sodium oxalate and magnesium oxalate

coprecipitate with calcium oxalate. A double precipitation of calcium

oxalate therefore is necessary. A considerable excess of ammonium oxalate

must be added because the magnesium present (and in some limestones

6Mellor, (Comprehensive Treatise on Inorganic and Theoretical Chemistry, Vol. XII,

Longmans, Green & Co., New York, 1932, pp. 221, 231.

Page 286: quimica inorganica cuantitativa


it is present in considerable amounts) forms complex oxalate salt?. The

contamination of calcium oxalate by magnesium oxalate is decreased

because of this complex ion formation and also because of the tendency

of magnesium oxalate to form supersaturated solutions. From the latter

fact it follows that prolonged digestion of the calcium oxalate precipitate

is undesirable, since it might lead to postprecipitation of magnesium


The precipitate of calcium oxalate may be treated in several ways

before weighing, (a) The precipitate, CaC 2 4.H2O, can be dried at 105

and weighed as such, though this is not recommended, (b) If calcium

oxalate is ignited at 500 25 it is converted quantitatively into the

carbonate. This procedure is recommended only if a muffle furnace in

which the temperature can be controlled very accurately is available.

(c) By treating the calcium oxalate with sulfuric acid it is converted into

oxalic acid, which may be titrated with standard potassium permanga-nate solution; from the results of the titration the calcium present in

the precipitate can be calculated (see p. 196). (d) The calcium oxalate

may be treated with sulfuric acid and the resulting calcium sulfate

weighed after expelling the excess acid. With proper care this method is

very satisfactory, (e) Finally, the calcium oxalate may be ignited over a

Meker burner or a blast lamp at which temperature it is converted into

the oxide (CaC 2O4 -> CaO + CO 2 + CO). This is perhaps the method

most often employed. Calcium oxide is hygroscopic and must be weighed

quickly in order to obtain a constant weight.

Procedure. The combined filtrates from the precipitation of the

hydrous oxides should be evaporated if necessary to a volume of about

200 ml. Add 2 drops of methyl red indicator and acidify the solution with

hydrochloric acid. Heat nearly to boiling and add a clear solution of 2 g.

of ammonium oxalate dissolved in 25 ml. of water. Put the beaker over a

low flame so as to keep the temperature of the solution at 70 to 90 and

add dropwise 2 N ammonium hydroxide from a buret mounted above the

beaker, stirring all the while. The addition of the base should take 10 or

15 minutes. When the solution is barely alkaline, set the beaker aside,

cover with a watch glass and let stand. As soon as the precipitate of

calcium oxalate has settled somewhat, test for completeness of precipi-

tation by adding 1 ml. of ammonium oxalate solution. After an hour

decant almost all of the supernatant liquid through a 7 cm. filter paperand wash the precipitate by decantation two or three times with cold

water, receiving the filtrate and washings in a 600 ml. beaker. (Add 50

ml. of concentrated nitric acid to the filtrate and place it on the steambath so that it may be evaporating while continuing the determination

of calcium.) Put the beaker containing the bulk of the precipitate underthe funnel and pour 25 ml. of 3 AT hydrochloric acid over the entire sur-

Page 287: quimica inorganica cuantitativa


face of the filter paper in order to dissolve the calcium oxalate on the

filter. Then wash the paper thoroughly with hot water and dilute the

filtrate to 250 ml. If the precipitate in the beaker has not completely

dissolved, heat the solution until it does. When almost to the boiling

point add about 0.5 g. of ammonium oxalate dissolved in 15 ml. of water

and then add dropwise, as in the first precipitation, 2 N ammonium hy-

droxide until slightly alkaline, as indicated by methyl red. Cover with a

watch glass and allow to stand for an hour.

(If calcium is to be determined volumetrically by titrating the acid

solution of the precipitate with permanganate, filter the calcium oxalate

through a Gooch crucible.) If the calcium is to be weighed in the form

of calcium oxide, filter the precipitate through an ashless filter paper.

After filtering wash thoroughly with a cold solution of ammonium oxalate,

5 g. to 1 1. of water. (Combine this filtrate and the washings with the

first filtrate and continue the evaporation already started.) Place the

paper containing the now clean precipitate in a weighed platinum crucible,

char the paper slowly, then heat at a moderate temperature over a Tirrill

burner and finally ignite at the full heat of a Meker burner or in a mufile

furnace at about 1100 if one is available. Cool in the desiccator for only

20 minutes, or until at room temperature, and, with the crucible cover

on, weigh as rapidly as possible (see p. 63). After constant weight is ob-

tained calculate the percentage of calcium oxide in the sample. Duplicate

results should not differ by more than 0.3 per cent.

Notes. Read p. 232 on the effect of the pll on the solubility of a salt like cal-

cium oxalate and then explain the precipitation of this salt in the above deter-

mination as the solution is made slightly alkaline with ammonium hydroxide.This method of bringing about the precipitation not only promotes a texture of

the precipitate which makes filtration easy, but also minimizes contamination

from coprecipitation.Calcium oxide, when exposed to air, not only will take up water but will com-

bine with carbon dioxide to form calcium carbonate as well; thus the necessity

for rapid weighing. It is so hygroscopic that if left overnight, for example, in a

desiccator charged with calcium chloride, it may even gain weight. Therefore it

should be left in such a desiccator only long enough to cool to room temperature.If it cannot be weighed immediately after cooling it should be kept in a desiccator

containing a desiccant like magnesium perchlorate. See the discussion of desic-

cants, p. 17.

DETERMINATION OF MAGNESIUM ^Magnesium is precipitated from an ammoniacal solution as mag-

nesium ammonium phosphate hexahydrate, which upon ignition is con-

verted to magnesium pyrophosphate./ The general properties of the

precipitate have been outlined under the discussion of the determination

of phosphorus/p. 256. Because the hexahydrate tends to occlude ammo-nium salts and alkali oxalates it is necessary to treat the combined

Page 288: quimica inorganica cuantitativa


filtrates from the calcium determination in a manner to eliminate or

destroy the ammonium and oxalate ions which the solution contains.

This is accomplished by evaporating the solution to dryness in the

presence of nitric acid.rOxalates are thus oxidized to carbon dioxide and

the ammonium ions to oxides of nitrogen and water.

Procedure. Add 50 ml. of concentrated nitric acid to the combined

filtrates from the calcium determination if this has not already been

done, and continue the evaporation to dryness. If a sizable residue re-

mains it is probable that considerable ammonium salts are still present;

in this case add a few milliliters of water, 25 ml. of concentrated nitric

acid and 10 rnl. of concentrated hydrochloric acid and repeat the evapo-

ration to dryness. Take care to avoid spattering during the last stages

of the evaporation. Then add 5 ml. of concentrated hydrochloric acid

and 25 ml. of water, warm and stir. A slight precipitate of silica, coming

largely from the solvent action of the several solutions upon glass vessels

throughout the analyses, may be present. Filter through a very small

filter paper, wash thoroughly with hot water and add the washings to

the filtrate. Dilute the filtrate to about 150 ml. and cool the solution,

preferably in ice. Dissolve 2 g. of diammonium hydrogen phosphate,

(NH 4) 2HP0 4 , in a small volume of water and add to the cold solution.

Put in 2 drops of methyl red and add dropwise concentrated ammonium

hydroxide until the indicator changes color; then add 5 ml. in excess.

Stir the solution for several minutes and allow to stand overnight or at

least 4 hours. .

Filter through a filter paper and wash thesprecipitate by dccantation

three or four times with cold 1:20 ammonium hydroxide. Discard the

washings. Place the beaker containing the major part of the precipitate

under the funnel and pour upon the filter 50 ml. of warm 1 : 10 hydro-chloric acid to dissolve the precipitate on the paper. Then wash, well with

1 : 100 hydrochloric acid. If the volume of the solution in the beaker is not

about 150 ml., dilute to that volume. Chill the solution and precipitate

magnesium ammonium phosphate hexahydrate exactly as before. Let

stand 4 hours or overnight and filter through an ashless filter paper.Then wash with 1:20 ammonium hydroxide until free from chlorides.

Ignite the precipitate in a weighed platinum or porcelain crucible in the

usual manner, using a Meker burner or a muffle furnace at 1000 to 1100.After obtaining a constant weight for the magnesium pyrophosphate cal-

culate the percentage of magnesium in the sample in terms of magnesiumoxide. The results for duplicate determinations should agree within 0.3

per cent.


Carbon dioxide is determined by treating a weighed sample of the

limestone with excess acid and absorbing the liberated gas in a suitable

Page 289: quimica inorganica cuantitativa


absorbing agent. The gain in weight of the vessel containing the absorb-

ing material represents the weight of the carbon dioxide in the sample.

Formerly a concentrated solution of potassium hydroxide was used to

absorb the carbon dioxide, but the usual practice at present is to employsolid materials such as Ascarite, which is a mixture of sodium hydroxide

and asbestos. It is obvious that the liberated carbon dioxide must be

dried before it is absorbed, else the gain in weight of the absorption

vessel would represent not the carbon dioxide alone, but moisture as

well. Provision must be made, furthermore, to trap the water formed

when the gas reacts with the sodium hydroxide during the absorption

(CO2 + 2NaOH -> Na 2C0 3 + H 20), or the gain in weight would be less

than the weight of the carbon dioxide. In addition, it is necessary to

prevent any other gas which might be liberated when the sample is

treated with acid from being taken up by the Ascarite. Hydrogen chloride

from the acid used and sometimes hydrogen sulfide must be trapped out.

All this necessitates a train of apparatus such as is shown in Fig. 36,

the details of which will be explained presently.

This experiment is typical of gas absorptions in general. Other com-mon examples are the determination of carbon and hydrogen in organic

compounds. It is not difficult to make accurate analyses by gas absorp-tion. It is imperative that the apparatus be gas-tight and that care be

taken to make accurate weighings of the absorption vessel.

Anhydrous calcium chloride often has been used to dry the gas before

it enters the absorption vessel. Others desiccants include: "Dehydrite,"

Mg(C10 4) 2.3H20; "Anhydrone," Mg(ClO4) 2 ; "Drierite," CaS0 4 ; and

"Dessichlora," Ba(C104)2. Anhydrous calcium chloride is not recom-

mended, for it frequently contains a small amount of calcium oxide which

would absorb carbon dioxide. If it is used to dry the gas before the liber-

ated carbon dioxide enters the absorption vessel it is necessary first to

treat it with carbon dioxide in order to convert any calcium oxide to

harmless calcium carbonate.

Apparatus. The apparatus is shown in Fig. 36. C is a wide-mouth

generator flask in which the sample is placed. It is fitted with a two-hole

rubber stopper. Inserted in the stopper is separatory funnel B which will

contain hydrochloric acid; also condenser D. In the mouth of the sepa-

ratory funnel a one-hole rubber stopper is placed and through it is

mounted a calcium chloride tube of Ascarite, A. The condenser is con-

nected through glass tube E to the train of U-tubes F, G and H. TheU-tube F contains anhydrous copper sulfate at the bottom of the tube

and a desiccant, say Anhydrone, in the top half of the two sides. Tube Gis charged with Anhydrone alone. Tube H is the absorption vessel; it

contains Ascarite for the most part but the side away from the generatorflask contains some Anhydrone as shown in the figure. Across the top of

Page 290: quimica inorganica cuantitativa



FIG. 36. Apparatus for the determination of carbon dioxide.

this tube a piece of copper wire may be attached to serve as a hook for

suspending the U-tube from the balance stirrup when weighing. Cotton

plugs should intervene between the two chemicals when more than one is

placed in a single tube; cotton also is inserted at the top in each arm of

the tubes to prevent dust from the absorbent or the desiccant from beingcarried along by the gas. Stoppers in the U-tubes should be corked tightly

and coated with sealing wax. Better still, U-tubes equipped with ground-

glass stopcocks may be used.

Tube H is connected to wide-mouth bottle K (containing a little

sulfuric acid) by a glass tube which in turn is coupled to a tube dipping

barely beneath the surface of the acid. The latter tube has an enlarged

portion which serves to prevent any acid from being drawn back into

tube H if the pressure in K happened to become greater than that in H.

Bottle K in turn is connected by glass tubing to bottle L which acts as a

trap for any water which might otherwise accidentally enter the train

from the aspirator (not shown in the drawing) which joins L through

heavy rubber tubing carrying pinchcock M. All units of the apparatusshould be connected glass to glass without unnecessarily long pieces of

rubber tubing between.

The apparatus is operated as follows: First it is tested for leaks.

Pinchcock N is closed and the aspirator is turned on and pinchcock M is

opened slightly. Bubbles will be seen in bottle K, but if the apparatus is

without leaks the bubbles soon will cease. If they continue all joints mustbe examined until the leak is found and repaired.

The several units of the apparatus serve the following purposes:

Separatory funnel B contains hydrochloric acid which may be intro-

Page 291: quimica inorganica cuantitativa


duced slowly through its stopcock to generator flask C containing the

sample. The resulting carbon dioxide passes into the condenser D which

returns any steam to C. Through E the gas goes into U-tube F where

water vapor is absorbed by the Anhydrone and any hydrogen chloride

or hydrogen sulfide is trapped out by the anhydrous copper sulfate.

Passing on to tube G, the gas is further dried by the Anhydrone in that

tube. Thus as the gas leaves G it is pure, dry carbon dioxide. Entering //

the carbon dioxide is absorbed by the Ascarite, and the water formed bythe reaction of the carbon dioxide and sodium hydroxide is caught and

retained by the Anhydrone in the top part of the right half of this tube.

Thus U-tube // gains in weight by an amount equal to the weight of the

carbon dioxide liberated from the sample. As already explained, bottle

K indicates the speed with which the gas is being drawn through the

apparatus; it also prevents any air from reaching tube // should the

pressure momentarily fluctuate in K. Bottle L catches any water which

might back-flow from the aspirator. The aspirator should not be turned

off while evacuating the apparatus without first closing M and opening

N, else water will be drawn back into the apparatus, though bottle Lwill trap it and prevent its going farther.

Procedure. When the apparatus has been proved gas-tight open

pinchcock TV, turn on the aspirator and regulate the sweep of air throughthe train by proper adjustment of pinchcock M. The rate should be such

that two or three bubbles per second pass through K. After 10 minutes

close M and turn off the stopcocks of U-tube H, or if it is not equippedwith stopcocks disconnect //, and quickly cap its small exit tubes with

short pieces of plugged rubber tubing or with policemen. Then weighU-tube H, handling it with a piece of Kleenex. Return H to the train,

sweep air through the apparatus for 10 minutes more and weigh U-tube

H again. Repeat this until constant weight is obtained (within 0.4 mg.).

Place an accurately weighed sample of about 1 g. of dry, finely

divided limestone in C and add about 25 ml. of water and several glass

beads. The water must cover the end of the upturned tube in C. Put

about 50 ml. of 1:1 hydrochloric acid in the separatory funnel. Dis-

connect the aspirator at K and slowly add acid from B through its

stopcock. Bubbles should be seen in .K" at a rate of two or three persecond. After all of the acid has been added to the flask, heat flask Cuntil the solution boils gently and continue heating until the evolution

of gas ceases. Turn off the flame, close pinchcock M, reconnect the aspi-

rator and open pinchcock M slowly so that air is drawn (pinchcock TV is

open, of course) at a slow rate through the train. After this has continued

for 30 minutes disconnect U-tube H and weigh it. The gain in weight

represents the carbon dioxide in the sample. Repeat the determination

with a new sample. Duplicate runs should agree within 0.3 per cent.

Page 292: quimica inorganica cuantitativa


Note. Errors in weighing the U-tube may result because of variations in the

amount of moisture on the glass surface of the tube. This error usually may be

avoided by using a similar but slightly lighter U-tube, which has been exposed to

the same atmosphere as has tube //, as a tare when making the weighing. If this

is done the weight of the carbon dioxide is equal, of course, to the number of grams

by which tube H exceeds its tare after the absorption, minus the number of grains

by which tube // exceeds the tare before the absorption. In other words, the

weight of the carbon dioxide is equal to the gain in weight of tube //.


Alloys or oxides containing lead usually may be dissolved with dilute

nitric acid, though the peroxide, Pb0 2 , and red lead, Pb 3 4 , require in

addition some hydrogen peroxide. After solution is effected evaporation

will expel the excess nitric acid, and lead sulfate may be precipitated

with sulfuric acid. Unless the nitric acid is driven off the solubility of

the lead sulfate would be considerably greater. In order to reduce further

the solubility of the precipitate, ethyl alcohol is added to the solution

before filtering. The lead sulfate is filtered through a Gooch crucible and

dried at 180 before weighing.

Procedure. Accurately weigh about 0.5 g. samples of the material

and transfer to 250 ml. beakers. Unless the substance is a soluble salt of

lead, add 50 ml. of 1:3 nitric acid and heat until the solution simmers.

After the reaction has ceased, if a brown residue is still undissolved add

3 ml. of 3 per cent hydrogen peroxide or more if necessary to bring about

complete solution. Add the peroxide slowly since a frothing will result.

When the solution is clear, cool and add 25 ml. of 1 : 1 sulfuric acid and

evaporate on a steam bath until the volume is approximately 15 ml.

Then heat over a low flame or on a hot plate until the first sulfuric acid

fumes appear. Cool, add 50 ml. of water, stir and add 75 ml. of 95 per

cent ethyl alcohol. Stir and put aside for 2 hours and then filter through a

Gooch crucible. Wash six or eight times by pouring 1:1 alcohol into the

crucible until two-thirds full. Apply suction only slightly both in filtering

and in washing. Dry the crucible and precipitate at 180 and weigh to a

constant weight. Calculate the percentage of lead in the sample. Dupli-

cate results should not differ by more than 0.2 per cent.


In the above determination of lead as lead sulfate, recourse wastaken to the use of an organic solvent to decrease the solubility of the

precipitate. Other organic solvents similarly employed are ether-alcohol

solutions and amyl alcohol. Organic precipitants are sometimes used to

precipitate inorganic ions. Usually such organometallic compounds have

quite small solubilities; as a rule they are distinctly and highly colored;

they almost always have high molecular weights so that small amounts

Page 293: quimica inorganica cuantitativa


of metal yield rather large amounts of precipitate; and because the

majority of organic precipitants are weak acids it is possible to commandconsiderable control over the precipitation of their metallic salts by

proper regulation of the pH. Though the advantage of these character-

istics has been utilized in various analyses, there are certain disadvan-

tages encountered. One is the fact that the organic precipitants often are

of rather low solubility, even in organic solvents, with the result that

there is danger of the precipitant itself separating out and contaminatingthe desired precipitate when an excess of the organic reagent is added.

Another is that the metallic salt of the organic compound frequently is

not wetted by water and this property causes the precipitate to creep

during filtration. The latter is more an annoyance than a real difficulty.

Furthermore, the action of the organic precipitants generally is not so

specific toward a given metal as might be desired.

Nevertheless the number of such precipitants which have been em-

ployed in inorganic analysis is quite large, as may be seen by consulting

special works on the subject.6 To illustrate those more frequently used

the following are briefly described: in addition, the procedures for the

determination of two metals, nickel and aluminum, by precipitation as

organometallic compounds will be given.


This reagent was used in the quantitative determination of nickel

by Brunck7 over 40 years ago. It is a weak acid having the formula

(CH 3C"NOII)2 which ionizes to yield one hydrogen ion. Divalent

nickel therefore reacts with two molecules of the compound to yield




a very insoluble red compound of which the nickel comprises 20.32 percent (56.69/288.8 = 0.2032). Dimethylglyoxirne often is referred to as a

specific precipitant for nickel. However, palladium also is quantitatively

precipitated by dimethylglyoxime, platinum is partly precipitated and

gold is reduced by it. The pH should be kept low, for cupric ion tends to

coprecipitate in alkaline solutions. With ores it is best as a rule to precipi-

tate all of the hydrogen sulfide group of metals, if present, before pre-

cipitating nickel with dimethylglyoxime. Aluminum, cobalt, chromium,iron, manganese and zinc do not interfere with the precipitation of nickel

6Sec, e.g., Prodinger, Organic Reagents Used in Quantitative Inorganic Analysis,

Elsevier Publishing Co., New York, 1940. Southern, J. Chem. Education, 18, 238(1941).

7Brunck, Z. angew. Chem., 20, 834 (1907).

Page 294: quimica inorganica cuantitativa


dimethylglyoxime under the conditions outlined below. The greatest value

of this method of determining nickel is that before this precipitant came

into use a dependable analysis of nickel alloys or ores containing cobalt

was practically impossible. The precipitant is added to a slightly acid

solution of the ore or alloy and then a small excess of ammonium hydroxide

causes the precipitate to appear. Tartaric or citric acid is added if metals

which arc precipit ated by ammonium hydroxide are thought to be present.

Determination of Nickel in Steel. Dimelhylglyoxime Method

Procedure. Accurately weigh out samples of steel turnings of about J

#. Transfer to 400 ml. beakers, add 50 ml. of 1:1 hydrochloric acid and

10 ml. of 1:1 nitric acid and warm until solution is complete. Allow to

simmer for several minutes until oxides of nitrogen are expelled. Then

dilute to 150 ml. and add 6 g. of citric acid. Slowly add concentrated

ammonium hydroxide until a piece of litmus paper touched lightly with

the stirring rod shows that the solution is barely alkaline, and then add

1 ml. in excess, ff necessary filter through a 7 cm. filter paper and wash

the paper thoroughly, receiving the washings in the beaker containing

the filtrate. Add 1:1 hydrochloric acid until the solution is just acid to

litmus, then boat to 60 or 70 and pour in 20 rnl. of a 1 per cent alcoholic

solution of dimethylglyoxime. Now slowly add ammonium hydroxide

until the solution is slightly basic and digest over a steam bath for an

hour. Filter through a previously weighed Gooch crucible. Test the fil-

trate for completeness of precipitation by adding 4 or 5 ml. of dimethyl-

glyoxime and digesting for 15 minutes. Wash the precipitate in the

crucible until free of chlorides. Dry in the oven at 105, cool in the

desiccator and weigh. Repeat, heating in the oven for 20-minute periods,

until a constant weigh is obtained. Calculate the percentage of nickel in

the steel. The percentage of nickel in many steels will run from a fraction

of 1 per cent to 2 or 3 per cent. Many stainless steels may contain as

much as 10 per cent of nickel or more. Duplicate results should differ bynot more than 0.05 per cent.

Notes. The pll of the solution must be carefully controlled. The nickel will

not precipitate if the solution is loo acidic, and any copper present will coprecipi-

tate if it is too alkaline.

If the steel is known to contain a high percentage of nickel the amount of the

precipitant needed for complete precipitation may exceed 20 ml. A simple calcu-

lation shows that 1 ml. of a 1 per cent solution of dimethylglyoxime will precipi-

tate 2.5 mg. of nickel. A 1 g. sample of steel containing about 10 per cent nickel

therefore would require approximately 40 ml. of precipitant. A large excess of

precipitant, however, must be avoided, first because some of the dimethyl-

glyoxime itself may precipitate from the aqueous solution, and secondly because

too much alcohol in the solution will increase the solubility of the nickel dimethyl-


Page 295: quimica inorganica cuantitativa


The tartaric or citric acid is necessary in order to convert the ferric iron into a

complex which is not precipitated by ammonium hydroxide. Citric acid is prefer-able because of the possibility of acid ammonium tartrate precipitating.

During the filtration the crucible should not be filled more than two-thirdsfull due to the tendency of the precipitate to creep.


Oxine acts as a precipitant for some two dozen metals, so that it

might seem to have little value in quantitative analysis. However, byusing buffers to control the pH of the solution certain separations canbe made. Its chief use has been in the determination of magnesium and

beryllium from ammoniacal solution. Many metals, including aluminum,may be precipitated with oxine from a buffered acetic acid solution. Fordetails of the use of this compound in quantitative separations one mayconsult the literature. 8 The use of oxine as the precipitant for aluminumis given below.

Determination of Aluminum. Oxine Method

(In the absence of ions other than alkalies, calcium, magnesium andberyllium)

Aluminum is usually determined by precipitation as hydrous alumi-num oxide in a manner similar to the gravimetric determination of iron.

The oxine method has the advantage of being more rapid. It may be

employed to isolate aluminum from magnesium, calcium and berylliumin solutions which are slightly acid. In basic solutions oxine separatesaluminum from phosphorus, arsenic and fluorine, and in ammoniacalsolutions containing hydrogen peroxide, from titanium and a number of

other metals.

Procedure. Obtain a soluble sample and accurately weigh an amountwhich contains 0.03 to 0.05 g. of aluminum. Dissolve in 100 ml. of water,warm the solution to 70 or 80, add 1 drop of concentrated hydrochloricacid and for each 3 mg. of aluminum thought to be present add 1 ml. of a5 per cent solution of 8-hydroxyquinoline dissolved in 2 TV acetic acid.

If a precipitate appears add 25 ml. of 2 TV ammonium acetate solution;if not, add the ammonium acetate solution until the first permanentprecipitate is obtained and then 25 ml. more. A yellow color in the solu-

tion indicates an excess of the precipitant. Let stand 1 hour and filter

through a weighed Gooch crucible. Wash with cold water until free ofchlorides and dry at 120. Weigh to a constant weight and calculate the

8 Koithoff and Sandell, J. Am. Chem. Soc., 50, 1900 (1928); Lundcll and KnowlesJ. Research Nail Bur. Standards, 3, 91 (1920); Hahn and Vieweg, Z. anal. Chem .

71, 122 (1927); Southern, J. Chem. Education, 18, 238 (1941); Hillebrand and Lundcll,Applied Inorganic Analysis, John Wiley & Sons, New York, 1929, p. 111.

Page 296: quimica inorganica cuantitativa


percentage of aluminum. The precipitate is A1(C 9H6ON)3. The average

deviation from the mean should not exceed 3 parts per 1000.



Cupferron was originally believed 9 to be specific for ion and copper

but it precipitates a number of metallic ions. The separation of iron,

titanium, and zirconium from manganese, aluminum or nickel furnishes

one example of the use of this precipitant.10 For an excellent summary

of the possibilities of cupferron in analytical work see the discussion given

by Hillebrand and Lundell. 11


This reagent precipitates a number of cations from slightly acidic

solutions, including cobalt, copper and iron, and separates them from

barium, calcium, strontium, manganese, nickel and zinc, among others.

Its chief application has been in the precipitation of small amounts of

cobalt in the presence of relatively high concentrations of nickel. Kales 12

warns that in cobalt analyses employing this reagent a blank always

should be run.


This compound will precipitate copper from slightly basic solutions.

It also forms an insoluble compound with hexavalent molybdenum in

acid solutions. It has been employed in the determination of copper in

the presence of molybdenum 13by precipitating the copper from ammo-

niacal solutions of steel containing both copper and molybdenum. Theresults are slightly high but are good, as checked against steels of known

copper content, and the analysis can be completed in 1 hour, whereas byother methods from 3 to 6 hours are usually necessary.


1. A 0.3000 g. sample of a chloride, when dissolved and treated with excess silver

nitrate, gives a precipitate weighing 0.5942 g. What is the percentage of chlo-

rine in the substance?

Answer: 49.00%.

2. A sample of potassium bromate weighing 0.3822 g. is reduced to potassiumbromide, after which an excess of silver nitrate solution is added. The precipi-

9 Baudisch, Chem. Ztg. t 33, 1298 (1909).w Brown, J. Am. Chem. Soc., 39, 2358 (1917).11 Hillebrand and Lundell, op. cil.

** See Fales and Kenny, Inorganic Quantitative Analysis, D. Appleton-Century Co.,New York, 1939, p. 349.

13 Kar, Ind. Eng. Chem., Anal. Kd., 7, 193 (1935).

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tate of silver bromide obtained weighs 0.4296 g. What is the percentage purityof the potassium bromate?

Answer: 99.95%.

3. A sample of 0.1000 g. of a pure oxide of iron is brought into solution and hy-drous ferric oxide is precipitated. The weight of Fe 2Os obtained is 0.1111 g.

Is the formula of the oxide FeO, Fe 2O 3 or Fe3O 4 P

4. A sample of iron ore weighing 1.000 g. is dissolved and the iron precipitatedas hydrous ferric oxide. After ignition the precipitate weighs 0.3850 g. Calcu-

late the percentage of iron in the ore in terms of (a) Fe, (b) Fe3O4 and (c)

FeS0 4.7H 20.

Answer: (a) 26.9%; (b) 37.2%; (c) 134.0%.

5. A sample of limestone weighing 1.208 g. yields a precipitate of magnesiumammonium phosphate which upon ignition is converted into Mg 2P207 weigh-ing 0.5020 g. Calculate the percentage of magnesium in the limestone in

terms of MgO.Answer: 15.05%.

6. The solution of a 1.050 g. sample of a silicate yields 0.2515 g. of Fe 2 3 andA1 2 3 . The oxides are dissolved and the iron is reduced to the ferrous state

after which it requires 26.72 nil. of 0.1000 N potassium permanganate solu-

tion for oxidation. Calculate the percentage of Fe 3 4 and A1 2O 3 in the silicate.

Answer: 19.61% Fe 3 4 ; 3.63% A1 2 3 .

7. (a) The calcium in 0.9082 g. of a substance is determined by precipitatingcalcium oxalate and igniting the latter to CaO weighing 0.5888 g. Calculate

the percentage present in terms of elementary calcium.

(b) If the calcium oxalate precipitate had been converted into CaSO 4 before

weighing, what weight would have been obtained ?

(c) If the calcium had been determined volumetrically by treating the cal-

cium oxalate with sulfuric acid and titrating with 0.1000 M potassium per-

manganate solution, how many milliliters of the permanganate would havebeen required?

Answer: (a) 46.34%; (b) 1.430 g.; (c) 42.01 ml.

8. A 1.000 g. sample consists entirely of sodium chloride and sodium bromide.It is dissolved and an excess of silver nitrate solution is added. The precipitateformed weighs 2.265 g. Calculate the percentage of NaCl and NaBr in themixture.

Answer: 70.2% NaCl; 29.8% NaBr.

9. A mixture weighing 0.5000 g. consists of NaCl and FeS04(NH4) 2S04.6H 20.The volume, in milliliters, of a barium chloride solution necessary to precipitatethe sulfur is exactly equal to the percentage of iron in the sample. Calculatethe number of grams per liter of BaCl 2.2H 2 in the barium chloride solution.

(In solving this problem it is necessary to look up the logarithm of two num-bers only.)

Answer: 43.74 g.

10. A sample of FeS0 4.7H 2 is dissolved and an excess of barium chloride is

added. The barium sulfate precipitated weighs 0.1400 g. If an identical

sample of FeS0 4.7H 2 is oxidized in acid solution with a potassium dichro-mate solution, 1.000 ml. of which is capable of oxidizing 0.006700 g. of sodiumoxalate, how many milliliters of the dichromate solution will be required ?

Answer: 6.00 ml.

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11. A sample consisting of a soluble chloride mixed with a soluble bromide is dis-

solved and an excess of silver nitrate is added. The resulting precipitate is

found to weigh 0.6032 g. The precipitate is treated with chlorine gas which

converts it completely into silver chloride. It is then found to weigh 0.4930

g. Calculate the percentage of silver in the original precipitate.

Answer: 61.52%.

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Chapter 16


JL REREQUISITES for analysis by titration methods in general are: the

rapid attainment of complete reaction between the standard solution andthat of the unknown ; the absence of side reactions which would nullify a

fixed stoichiometric relationship between the two reactants; the abruptappearance at or quite near the stoichiometric point of some propertywhich visibly will mark the point of chemical equivalence for the re-

actantsl In the case of reactions in which a precipitate is one of the

products the completeness and the reaction velocity are satisfactory, and

usually only one reaction takes place so that there exists a definite andknown stoichiometric relationship. Furthermore, the precipitation of anion as one of its insoluble compounds obviously must promote a great

change in the concentration of that ion in the supernatant liquid after

the stoichiometric point has been reached. However, the precipitationreactions which are adapted to a feasible indication of the stoichio-

metric point are relatively few. Those which are suitable may be classified

into four groups.' The detection of the end point for these precipitationtitrations is made possible by the development immediately after the

stoichiometric point of (1) a colored secondary precipitate, e.g., Mohr'smethod for chlorides; \2) a turbidity, e.g., Liebigs' cyanide method;(3) a colored soluble compound, e.g., Volhard's method for chlorides,bromides or iodides; and (4) the adsorption of certain organic dyes bythose particles of the precipitate in a colloidal state, e.g., Fajail's methodfor determining halides.j These four classes will now be discussed andactual analyses embodying the principle of each will be given.


The Mohr method for the determination of chloride by titration witha standard solution of silver nitrate illustrates the use of a secondarycolored precipitate to mark the end point. Only a very small concentra-tion of chloride ion may be at equilibrium with silver ion which simul-

taneously is at equilibrium with chromate ion (see p. 234). If, therefore,a solution of a chromate is added to an unknown solution of chloride andthe resulting solution is titrated with a silver nitrate solution, precipitation


Page 300: quimica inorganica cuantitativa


of pure silver chloride takes place until the chloride ion concentration is

reduced to a negligibly small value. Further addition of silver ion causes

silver chromate to precipitate and the deep red color of that compoundindicates the end point of the titration./

The concentration of chromate ion necessary to bring about the pre-

cipitation of the red silver chromate at the stoichiometric point for silver

chloride is calculable. The KB .V .value for silver chloride is about 1 X 10~ 10

and for silver chromate is 2 X 10~ 12. When chloride is titrated with silver

nitrate solution the concentration of the silver ion at the stoichiometric

point is \/iO~ 10 or 10~ 5 moles per liter. When the silver ion concentration

has this value the concentration of chromate ion necessary to precipitate

the red silver chromate must slightly exceed 0.02 M because

- -02

However, if the unknown chloride solution is made 0.02 M with respect

to chromate, the solution becomes colored so deeply yellow that the

formation of a small amount of silver chromate immediately following

the stoichiometric point is obscured; it is customary therefore to make

the concentration of the chromate ion only about 0.003 M. Its concen-

tration must not be much lower than this, however, for it is obvious that

the smaller its value the greater must be that of the silver ion in order

that Knt>for silver chromate be exceeded. Suppose, for example, that

the solution of the chloride were made 0.0001 M with respect to the

chromate ion. The concentration of silver ion necessary to make the solu-

tion saturated with silver chromate would be

2 v 10~ 12

[Ag+P = !T- =2X10-*

[Ag+]= 1.4 X 10- 4

Now if the volume at the end of the titration were 100 ml. and if the

silver nitrate solution were 0.1 M, the volume of excess silver nitrate

needed to the point of silver chromate precipitation would be 0.14 ml.

for if X be the excess volume of silver nitrate solution, in milliliters,

A(o.l) = 1.4 X 10-4

X = 0.14 ml.

A similar calculation reveals that if the chromate indicator of a concen-

tration of 0.003 M is used, the volume of 0.1 M silver nitrate solution

needed to make the solution saturated with respect to silver chromate is

only 0.026 ml. This is hardly a significant error, whereas 0.14 ml. is.

Page 301: quimica inorganica cuantitativa


The calculation, which the student should confirm, also shows that this

excess of 0.026 ml. of 0.1 M silver nitrate establishes a silver ion con-

centration of 2.6 X 10~ 5which, it will be noted, is slightly greater than

that which exists at the stoichiometric point for the silver chloride equi-

librium, namely 10~6. Thus a slight positive titration error is inevitable

in the Mohr method; that is to say, the volume of the standard silver

nitrate solution necessary for the end point must be slightly greater than

that corresponding to the stoichiometric point. Exactly how much greater

is computed as follows for the example cited: At the end point (wheresilver chromate begins to precipitate) [Ag

+]= 2.6 X 10~~ 6 moles per liter.

At the silver chloride stoichiometric point lAg1

"] ^ 10~6 moles per liter.

The difference, 1.6 X 10~6 moles per liter, or 1.6 X 10~6 moles per 100

ml., represents the titration error. Since 1000 ml. (of 0.1 M silver nitrate)

contains 0.1 mole, then 0.016 ml. contains 1.6 X 10~6 moles. The titra-

tion error, then, is 0.016 = 0.02 ml. of silver nitrate solution. Of course

the titration error is eliminated in practice if the silver nitrate solution is

standardized under conditions identical with those of the analysis.

Because the concentration of the chromate indicator ion plays such

an important role in the titration error, the acidity of the solution beingtitrated must be carefully regulated. The ionization constant of chromic

acid is quite small; furthermore, in the slightly acidified solution the

chromium will exist largely as the dichromate ion. For this reason a pll

lower than 6 is unsuitable; and a pH greater than 10 causes the precipi-

tation of silver oxide. The custom therefore is to adjust the pH of the

titrated solution of chloride between these values.


Procedure. Preparation and Standardization of Silver Nitrate

Solution. Weigh out approximately 8.5 g. of pure silver nitrate and dis-

solve in 500 ml. of chloride-free water. (See second paragraph under

Notes, p. 250.) Keep in a brown, glass-stoppered bottle. This solution is

standardized using pure sodium chloride as the standard. Heat several

grams of finely divided, pure sodium chloride in a porcelain crucible for

15 minutes at a dull red heat. Place while still warm in a glass-stopperedbottle and store at once in a desiccator. When cool, accurately weighabout 0.2 g., transfer to a 250 ml. porcelain casserole, dissolve in 50 ml.

of chloride-free water and add 2 ml. of 0.1 M potassium chromate solu-

tion. Titrate with the silver nitrate solution. Although silver chromate

is red, and although as the end point is approached there will be a mo-

mentary flash of red color which disappears, the end point is not marked

by a true red color. Up to the end point the white precipitate of silver

chloride and the yellow solution of potassium chromate combine to give a

bright lemon-peel color. But at the end point the presence of a slight

Page 302: quimica inorganica cuantitativa


precipitate of red silver chromate blends with the yellow to give a rather

muddy yellow color. This is the end point. From the volume of silver

nitrate solution and the weight of sodium chloride calculate the normalityof the solution, using equation (1), p. 81, as usual. Duplicate determi-

nations should agree within 1 part per 1000.

Procedure for Analysis. Accurately weigh about 0.2 g. samples of

the chloride unknown, transfer to 250 ml. porcelain casserole, dissolve in

50 ml. of chloride-free water and titrate with the standard silver nitrate

solution exactly as in the standardization procedure, but only after test-

ing for acidity. To do this take a small amount of the material and dis-

solve in half a test tube of water. If litmus shows the solution to be

acid, add 0. 1 N sodium hydroxide to the solution in the casserole to the

phenolphthalein color change; then add a drop or two of very dilute

nitric acid to destroy the red color. If the test shows that the sample

gives an alkaline solution, add very dilute nitric acid to the solution in

the casserole until the red color of phenolphthalein barely turns colorless.

Calculate the percentage of chloride in the sample. The deviation from

the mean for duplicate determinations should not exceed 1 part per 1000.

Notes. The Mohr method can be applied to the determination of bromide, in

the absence of chloride. Neither of these anions can be determined in the

presence of other anions, such as phosphate or sulfide, which form insoluble

silver salts in neutral solution. The method cannot be used in the determination

of iodide or thiocyanate because the end point is not sharp, due to adsorptive

properties of the precipitate.Problem. 32.10 ml. of a silver nitrate solution is required in the titration of

0.1752 g. of pure sodium chloride. Calculate the normality of the solution. Whatis the "sodium chloride value" of the solution i.e., the weight of sodium chloride

equivalent to 1 ml. of the solution?

Solution. From equation (1), p. 81, we have

N.CI value .-

, 0.0054581000

Problem. A chloride unknown weighing 0.2012 g. is titrated by the Mohrmethod, 25.50 ml. of 0.0934 N silver nitrate solution being required. Calculate

the percentage chloride in the sample in terms of sodium chloride.

Solution. % Nad = .



i The examples of this type of titration are relatively few; the Liebigmethod for cyanides is an important example. The method involves the

formation of a soluble and stable complex as the standard solution reacts

Page 303: quimica inorganica cuantitativa


with the unknown; after the stoichioinetric point is reached there is a

sudden increase in the concentration of the ion being added from the

buret, and this leads to the formation of the difficultly soluble substance

which causes the turbidity. To illustrate: when a cyanide solution is

titrated with silver nitrate the reaction up to the stoichiometric point is

2CN- + Ag+-Ag(CN) 2-

but with a slight excess of silver ion a precipitate (turbidity) results due

to the reaction

Ag(CN),- + Ag+ -> Ag [Ag(CN) 2 ]J(Review these equilibria as given under Complex Ions and Solubility, p. 231.)

Obviously the titration error will be positive since the eye may fail to

detect the first faint turbidity. However, a negative titration error is

possible if, due to insufficient stirring during the titration, the silver ion

concentration becomes locally so great as to form some precipitate before

the stoichiometric point is actually reached. The end point is more readily

detected by adding to the cyanide solution before starting the titration

some ammonium hydroxide and a small quantity of potassium iodide.

Under these conditions the silver argenticyanide does not precipitate

after the stoichiometric point has been reached due to the formation of

the silver ammonia complex, Ag(NH3) 2+

, but, instead, a turbidity results

because of the precipitation of silver iodide:

Ag(NH 3) 2+ + I- -> Agl + 2NH 3

In working with cyanides their great toxicity should be kept constantly

in mind. Not only should the hands be washed thoroughly after handling

cyanides, but one should avoid carelessly pouring cyanides into the sink

without flushing down the drain. Any acid which might be in the sink

would react with the cyanide to liberate the extremely poisonous hydro-

gen cyanide gas.


Procedure. Accurately weigh samples of about 0.3 to 0.5 g. and

transfer to 250 ml. Erlenmeyer flasks. Dissolve in 25 ml. of water, add

5 ml. of concentrated ammonium hydroxide and 0.1 to 0.2 g. of potassiumiodide. Dilute to 100 ml. and titrate with approximately 0.1 M silver

nitrate solution. Calculate the percentage of cyanide in the sample.

Duplicate determinations should show a deviation from the mean under

2 parts per 1000.

Problem. A cyanide sample weighing 0.5206 g. is titrated with a 0.1000 Msilver nitrate solution, 25.08 ml. being required. Calculate the percentage of

cyanide present in terms of sodium cyanide.

Page 304: quimica inorganica cuantitativa


Solution. The normality of the silver nitrate solution must be regarded as

twice as great when reacting with cyanide as when reacting with chloride because

one silver ion reacts with two cyanide ions but with only one chloride ion. There-

fore this silver nitrate solution must be regarded as being 0.2000 N for the prob-

lem at hand. The "sodium cyanide value" of the solution (i.e., the weight in

grams of sodium cyanide equivalent to 1 ml. of the silver nitrate solution) is,


^ = 0.0098041UUU


% NaCN = -


The Volhard method for chlorides is the most common example of a

precipitation titration in which a colored soluble substance, formed when

the stoichiomctric point is reached, marks the end point. The method is

satisfactory also for the determination of bromide or iodide. In the

analysis for the halides an indirect approach is employed in the sense

that an amount of standard silver nitrate solution in excess of that re-

quired by the halide is added, and the excess of silver ion is determined

by titrating with a standard thiocyanate solution. Silver may be deter-

mined also by this method but a direct titration of the unknown silver

solution is made with a standard thiocyanate solution. (This is the pro-

cedure in standardizing the latter solution.) In cither the direct or the

indirect approach the reactions which take place, in order, are

Ag+ + SCN- -> AgSCN (white)

andFe+++ + 6SCN- - Fe(SCN)r (red)

The latter reaction is that of the indicator which is added to the solu-

tion being titrated in the form of ferric ammonium sulfate. The white

silver thiocyanate precipitates and when the stoichiometric point of this

reaction has been reached the next drop of thiocyanate solution brings

about the formation of the red ferric thiocyanate complex ion.

The titration is carried out in nitric acid solution in contrast to other

volumetric methods for silver or halide determinations which require a

neutral solution. The metals of group 2 cause no serious trouble and, in

the presence of nitric acid, such anions as carbonates, oxalates and phos-

phates, which would precipitate as the silver salts in neutral solution,

do not interfere.

In the analysis of a chloride it is necessary to filter off the silver

chloride whic"h is precipitated after the addition of excess standard silver

nitrate solution. This is due to the fact that if the solution is titrated

Page 305: quimica inorganica cuantitativa


with standard thiocyanate without first being filtered, the thiocyanatewill react not only with the excess silver ion which was added, but, in

addition, with some of the silver chloride. In analyses for bromide or

iodide, on the contrary, the silver halide need not be filtered off because

silver bromide and silver iodide are so very insoluble that no appreciablereaction takes place between these precipitates arid the thiocyanatesolution.


Procedure. Preparation and Standardization of Solutions. Pre-

pare an approximately 0.1 M solution of silver nitrate following the in-

structions given under the Mohr method, p. 283. Prepare a solution of

approximately 0.1 M potassium thiocyanate by weighing roughly 9.7 g.

of the salt and dissolving in a liter of water; store in a glass-stopperedbottle. Prepare the indicator solution by saturating 100 ml. of 1 N nitric

acid with ferric ammonium sulfate.

Determine the relative concentrations of the two standard solutions

as follows: Fill a buret with the silver nitrate solution and run some40 ml. into a 250 ml. Krlenmcycr flask. Read the buret. Add 5 ml. of

1:1 nitric acid which has been freshly boiled and cooled, and 1 ml. of

indicator solution. From a second buret filled with the thiocyanate solu-

tion titrate until a reddish brown color is obtained. Approach the end

point slowly and with vigorous agitation between the dropwise additions

from the buret. From the buret readings calculate how many milliliters

of silver nitrate solution are equivalent to 1.000 ml. of the thiocyanatesolution.

Standardize the thiocyanate solution in the following manner: Accu-

rately weigh approximately 0.2 g. samples of pure, dry sodium chloride,

transfer to 250 ml. Erlenmeyer flasks and dissolve in 50 ml. of chloride-

free water. Add 5 ml. of 1:1 nitric acid and then add from a buret the

silver nitrate solution already prepared until a definite excess of 5 ml.

or more is present. (When an excess has been added the precipitate of

silver chloride will coagulate arid settle quickly. Allow to settle, then addmore silver nitrate from the buret; if no further precipitation takes placein the supernatant liquid an excess is assured.) Filter through a Goochcrucible and wash well. To the filtrate arid washings add 1 ml. of the

indicator solution and titrate with the thiocyanate solution to the reddish

brown color. Knowing the relative strengths of the silver nitrate and

thiocyanate solutions, calculate the volume of silver nitrate which wasadded in excess (i.e., that which did not react with the chloride). Sub-tract this volume from the total volume of silver nitrate originally added.The difference is the volume of silver nitrate equivalent to the sodium

chloride; from this calculate the normality of the silver nitrate solution.

Page 306: quimica inorganica cuantitativa


The deviation from the mean for duplicate determinations should not

exceed 1 part per 1000.

Problem. A sample of 0.1302 g. of pure, dry sodium chloride is dissolved and

24.82 ml. of 0.1000 N silver nitrate solution is added. The solution then is titrated

with potassium thiocyanate solution, of which 1.000 ml. is equivalent to 1.092 ml.

of the silver nitrate solution. The titration requires 3.33 ml. of thiocyanate solu-

tion. Calculate the normality of the silver nitrate solution.

Solution. Since 1.000 ml. of thiocyanate solution is equivalent to 1.092 ml. of

silver nitrate solution, 3.33 ml. of the former is equivalent to 3.64 ml. of the latter.

Therefore 24.82 3.64 = 21.18 ml. is the volume of silver nitrate which reacted

with the 0.1302 g. of the standard, NaCl. Then from equation (1), p. 81,

* - -

Procedure for Analysis. Accurately weigh samples of a soluble

chloride amounting to about 0.2 g. Transfer to Erlenmeyer flasks and

proceed with the analysis following the directions given just above for

the standardization of the silver nitrate solution by the Volhard method.

Since the normality of the silver nitrate solution is known, the percentageof chloride in the sample may be calculated. The deviation from the meanshould be 1 part per 1000 or less.


The dispersed phase of a colloidal system, because of the large surface

area", usually shows a high adsorptive power. Thus, if a chemical reaction

occurs in which an insoluble compound which partly assumes a colloidal

state is formed, the particles are found to adsorb ions from the solution

which constitutes the dispersion medium. The ion which is adsorbed is

that ion of the precipitate itself which is the more abundant in the solu-

tion. A colloidal dispersion of silver chloride, for example, will adsorb

either silver ions or chloride ions, depending upon which of the two is

present in greater concentration in the solution. After a layer of ions hasbeen adsorbed by the colloidal particle there follows a secondary adsorp-tion consisting of a layer of ions of the opposite charge to that of the

primary layer.

This, in brief, is the theory which forms the basis of adsorption indi-

cators for detecting the end point in certain precipitation titrations. Ofthe two ions which form the precipitate there obviously will be in the

surrounding solution an excess of one of them up to the stoichiometric

point, but an excess of the other afterward. For example, if a chloride

solution is titrated with silver nitrate solution, silver chloride precipi-tates (and some silver chloride is colloidally dispersed) as fast as silver

ion is added from the buret; thus the chloride ion concentration exceedsthat of the silver ion up to the stoichiometric point. It follows that under

Page 307: quimica inorganica cuantitativa


these conditions the ion primarily adsorbed will be the chloride ion and

the secondary layer will consist of any positive ion present, e.g., sodium.

As the stoichiometric point is approached the concentration of the chlo-

ride ion in solution becomes progressively smaller; at the stoichiometric

point the concentrations of the chloride ion and the silver ion become

equal. Immediately past the stoichiometric point the silver ion predomi-nates and therefore the colloidal particles now suddenly adsorb silver ions

as the primary layer.

Since the adsorbent has the tendency preferentially to adsorb one of

its own ions from the surrounding solution as a primary layer, an organic

dye like fluorescein may be added to the solution without the adsorption

of the negative ion of fluorescein. Therefore when a chloride solution

which also contains this indicator is titrated with silver ion, the precipi-

tated silver chloride remains white up to the stoichiometric point. Just

past the stoichiometric point the silver ion is primarily adsorbed; thus

now the secondarily adsorbed layer must be of a negative sign and the

negative ion of fluorescein is adsorbed as the secondary layer. Moreover,the dye which, in solution, had a greenish yellow color, shows a pinkcolor when adsorbed by the silver chloride particles. The change of the

white silver chloride to a precipitate superficially pink in color thus marks

the end point in the titration. ^

To summarize: up to the stoichiometric point silver chloride particles

primarily adsorb chloride ions in preference to the negative ion of fluo-

rescein; the secondary layer consists of some positive ion of the solution.

The precipitate has its normal white color. With the addition of a slight

excess of silver ions the particles primarily adsorb silver ions and the

particles become positively charged. By secondary adsorption they then

attract the negative fluorescein ions which cause the pink coloring.

In general adsorption indicators may be successfully employed only

if certain conditions are fulfilled. The precipitate which is formed must

be partly dispersed as colloidal particles so that the adsorptive forces

will be great. If the tendency of the precipitate toward coagulation is

too great (as happens in the presence of large amounts of neutral salts)

it may be necessary to employ a protective colloid e.g., dextrin to

maintain the dispersion. The indicator ion must bear a charge of the same

sign as that of the titrated ion, and (a) it must not be preferentially

adsorbed as a primary layer before the stoichiometric point is reached,

but (b) it must be preferentially adsorbed (i.e., rather than any other

ion of the same sign like the nitrate ion in the chloride-silver titration)

as a secondary layer immediately after the stoichiometric point is reached.

It must be noted that fluorescein is a very weak acid, and since the

color change is due to the adsorption of its negative ion, it follows that

anything which renders the concentration of that ion negligibly small

Page 308: quimica inorganica cuantitativa


will impair or even destroy its capacity as an adsorption indicator. The

hydrogen ion concentration therefore must be low; in a solution having a

pH of 7 to 10 fluorescein functions very well. Eosin and dichlorofluo-

rescein are stronger acids than fluorescein and may be used in acid solu-

tions. However, the former cannot be used in the titration of chlorides

by silver ion because its negative ion is so strongly adsorbed that the

primary layer formed during the approach to the stoichiometric point

does not consist exclusively of chloride ions. Thus a reddish color begins

to appear before the stoichiometric point is reached.

For the application of other dyes as adsorption indicators see Kolthoff

and Stenger, Volumetric Analysis, Vol. II, Interscience Publishers, New

York, 1947.


Procedure for Standardization. Prepare a solution of silver nitrate

about 0.1 M in the manner given for the Mohr method for determining

chlorides. Accurately weigh samples of the standard, sodium chloride,

which has been heat-treated as described under the Mohr method, of

about 0.2 g. each, transfer to 250 ml. Erlenmeyer flasks and dissolve in

about 50 ml. of chloride-free water. Prepare the indicator solution by

dissolving 0.1 g. of fluorescein in 100 ml. of 70 per cent alcohol or 0.1 g.

of sodium fluoresceinate in 100 ml. of water. Or dissolve 0.1 g. of dichloro-

fluoresccin in 100 ml. of 70 per cent alcohol or 0.1 g. of its sodium salt in

100 ml. of water. Add 10 or 12 drops of cither indicator to the sodium

chloride solution and 0.1 g. of dextrin. Titrate, with continuous agitation,

with the silver nitrate solution. Keep away from direct sunlight. The end

point has been reached when the white precipitate suddenly takes on a

reddish tinge. From the volume of silver nitrate solution used and the

weight of the standard employed, calculate the normality of the solution

and its chloride value. Duplicate results should not deviate from the

mean by more than 1 part per 1000.

Procedure for Chloride Analysis. Accurately weigh out samples of

0.2 g. or slightly more of the dried material for analysis. Transfer to

250 ml. Erlenmeyer flasks, dissolve in about 50 ml. of chloride-free water

and proceed according to the above directions for the standardization.

From the normality or the chloride value of the silver nitrate solution

and the volume usefd for the titration, calculate the percentage of chloride

in the' sample. Duplicate determinations should show a deviation from

the mean not exceeding 1 part per 1000 for samples of fairly high chloride


Notes. Dextrin retards the coagulation of silver chloride and thus maintains

the adsorptive property of the particles.

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The above method may be applied to the titration of bromides, iodides and

thiocyanates as well as chlorides. Eosia may be used as the indicator in the analy-sis of iodides or thiocyanates if the pll of the solution is greater than 3.

Spot Test Indicators. In certain titrations for which no internal

indicator has been discovered a drop of the titrated solution is removed

as the stoichiometric point is approached. This drop i3 tested on a porce-lain spot-plate with a drop of a reagent which will reveal the presenceof the titrated substance or of the titrating solution. The end point for

the titration has been reached when a spot test shows the absence of the

constituent being titrated. For an example of this method of determiningthe end point see pp. 202 and 203.

Questions and Problems

1. Show by calculations that 2 ml. of 0.1 M potassium chromate solution will

establish a suitable concentration of chromate indicator ion in a Mohr titra-

tion if the final volume is about 75 ml.

2. Explain why, in the determination of chloride by the Mohr method, the pHof the solution should be adjusted to a value between 7 and 10.

3. Hydrochloric acid is, of course, a strong acid. Would it be possible, neverthe-

less, to standardize a hydrochloric acid solution by a Mohr titration? Explain.

4. Calculate the titration error when 50 ml. of an approximately 0.1 AT solution

of a chloride is titrated with 0.1 N silver nitrate solution, using 10~ 5 M potas-sium chromate as the indicator.

5. If M represents the molarity and N the normality of a silver nitrate solution

explain why M = N when the solution is used in a Mohr titration, whereas2M = N when the solution is used in a cyanide determination by the Liebigmethod.

6. In the Volhard method why must the precipitate of silver chloride be filtered

before the solution may be titrated with standard thiocyanate solution,

whereas silver bromide or silver iodide need not be filtered off before the


7. In the chloride determination by the adsorption indicator method using fluo-

rescein as the indicator, why must the pH of the solution be above 7? Whybelow 10? Why may dichlorofluorescein be used in an acid solution of a

chloride? Why is eosin not suitable as an adsorption indicator in a chloride


8. What is the molarity of a silver nitrate solution if 33.86 ml. of it is required to

titrate a 0.4902 g. sample of pure, dry potassium cyanide?Answer: 0.1112 M.

9. A cyanide sample weighing 0.5100 g. is titrated with 0.1010 M silver nitrate

solution, 24.48 ml. being required. Calculate the percentage of cyanide pres-ent in terms of sodium cyanide.

Answer: 47.5%.

10, A silver nitrate solution is available 1.000 ml. of which is equivalent to

0.007508 g. of potassium chloride. A 0.2452 g. sample of a chloride is analyzed

Page 310: quimica inorganica cuantitativa


by the Volhard method. A volume of 28.00 ml. of the silver nitrate solution

is added and the solution is titrated with a potassium thiocyanate solution

1.000 ml. of which is equivalent to 1.046 ml. of the silver nitrate solution.

The titration requires 5.15 ml. of thiocyanate solution. Calculate the per-

centage of chloride in the sample.Answer: 32.9%.

Page 311: quimica inorganica cuantitativa

Chapter 17


IT HAS been mentioned already in Chapter 10 that, inasmuch as it is

possible to calculate the e.rn.f. developed at the stoichiometric point by a

given cell, the end point for a titration may be detected without the use

of color indicators. It is only necessary that the e.m.f. at the stoichio-

metric point be computed and the titration carried out until this e.m.f.

is registered. It is common practice to make numerous readings of the

e.m.f. of the cell during a titration and to carry the titration somewhat

beyond the stoichiometric point. Since the e.m.f. varies sharply in the

vicinity of the stoichiometric point, it is not necessary to take manyreadings until near this point, but here the recordings should be madeafter the addition of each drop from the buret. After the data have been

secured a graph may be constructed, plotting the e.m.f. as the ordinate

against volume of reagent added as the abscissa. A perpendicular droppedfrom the steepest portion of the curve will indicate the volume of reagent

necessary for the titration.

The Potentiometer. Potentiornetric titratioiis are carried out by use

of the instrument called the potentiometer. Fig. 37 illustrates a simple

type of potentiometer connected to the indicator electrode (in the solu-

tion to be titrated) and a reference, calomel electrode. The purpose is to

measure the e.m.f. of the cell during the course of the titration. This

cannot be done with an ordinary voltmeter since the current necessary

to deflect the pointer, coming as it does from the cell, causes changes in

concentration within the cell, thereby diminishing the e.m.f. even while

the measurement is being made. To circumvent this situation the e.m.f.

of the cell, Ex in the figure, is balanced with an opposing e.m.f. from Ba

so that no current flows from the cell.

A wire BA of uniform resistance is connected by heavy copper wire

of no appreciable resistance to battery Ba which usually is a lead storage

battery but may be two dry cells. Ex is the cell the e.m.f. of which is to

be measured. It is connected, on the one side, through galvanometer G,

to the resistance wire at terminal ZJ, the zero point of resistance. On the

other side Ex is connected, through tapping key K, to sliding contact Cwhich may be moved along BA to such a position that the key may be

1 For a review with a bibliography see Furman, J. Ind. Eng. Chem., Anal. Ed., 14,

367 (1942).


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tapped without causing any deflection of the galvanometer. The galva-

nometer thus functions as a nullpoint instrument. When this condition

prevails the e.m.f. from Ba is just balanced by the opposing (note the

two polarities) e.m.f. from Ex . The ratio of the two electromotive forces

must now be equal to the ratio of the corresponding resistances, or

Ex :Ba = BC:BA, and, since the resistances are directly proportional

to the lengths of the resistance wire involved, it follows that Ex:Ba =

(length of BC): (length of BA). Of these four terms Ba and tho length

BA are known, and the length BC is measured; thus Ex is readily


FIG. 37. Principle of the potentiometer as used in potentiometric titrations.

(Ba) Battery. (R) Variable resistance. (BA) Slide wire. (C) Sliding contact. (K)

Tapping key. (/) Indicator electrode. (Cal) Calomel electrode. (G) Galvanometer.

(Bu) Buret. (Ex) The cell the e.m.f. of which is to be measured.

Page 313: quimica inorganica cuantitativa




FIG. 38. Apparatus for precise potentiometric titrations. (Courtesy, Popoff:

Quantitative Analysis, Philadelphia, The Blakiston Company.)

The e.m.f. from Ba is made known by replacing Ex of Fig. 37 (e.m.f.

of Fig. 38) with a standard cell which provides an accurately knowne.m.f. The Weston cell with an e.m.f. at 20 of 1.0183 volts is commonlyemployed for this purpose. The e.m.f. from Ba may be regulated to anydesired value by means of the rheostat /?.

In potentiometric titrations the absolute values of Ex are not neces-

sary; it is sufficient to read the changes in BC, particularly in the

neighborhood of the stoichiometric point, from the instrument. Since

BA is constant, K is directly proportional to BC and therefore the

readings of BC indicate the change of potential in the cell, Ex .


The Hydrogen Electrode. Obviously potentiometric titrations maybe made in the titration of acids and bases provided a relation exists be-

tween the hydrogen ion concentration and the electrode potential of a

given electrode immersed in the solution being titrated. The hydrogenelectrode may be so used. If the hydrogen electrode dips into the unknownsolution and is connected to another half-cell, for example to a calomel

half-cell as shown in Fig. 39, an electric current will flow in the external

circuit. An instrument for measuring the e.m.f. must, of course, be inserted

in the external circuit. This cell may be represented


KC1 (1 Af)H+(XM)Hz (1 atm.)


Page 314: quimica inorganica cuantitativa


From Table 17 we have

Hg2Cl 2 + 2e = 2Hg2H+ + 2e = H 2,

2C1-, E Cal= 0.28

E H = 0.00

so that the e.m.f. of the cell, Ec ,= EI E 2 or


E. == E Cai- K H - 0.0591 log [H+]

= 0.28 - 0.00 - 0.0591 log [H+]

E = 0.28 - 0.0591 log [H+]

FIG. 39. Calomel and hydrogen electrodes for a

potentiometric titration.

Since pH =log [11+] we may substitute in equation (1) and obtain

Kr= 0.28 + 0.0591 pH

from whence

(2) pH = EC - 0.28


The hydrogen electrode is subject to poisoning and, even with greatcare in handling it, erroneous readings may be obtained or the circuit

actually may go dead. As a result it is usual to employ some other

electrode the potential of which also is proportional to the hydrogen ion

concentration. One of the most commonly used is the glass electrode.

Page 315: quimica inorganica cuantitativa


The Glass Electrode. 2 The glass electrode consists of a glass tube

terminating in a thin glass bulb containing a solution of hydrochloric

acid, 0.1 N or less. An electrode extends through the tube and dips into

this solution. The electrode must have a constant potential in regard to

the hydrochloric acid solution. It may be a silver electrode coated with

silver chloride or it may be a platinum electrode; if the latter, then some

quinhydrone (see next section) is added to the solution, making it a

quinhydrone electrode. The glass bulb is placed in the solution the pH of

which is to be determined. The bulb acts as a membrane which would

seem to be permeable to hydrogen ions. If a calomel electrode also is

placed in the unknown solution the e.m.f. between the calomel and the

glass electrodes may be measured. The cell may be represented thus:


HC1 (0.1 M) GlassSolution being I! Ka (1analyzed


(Reference electrode)

The e.m.f. established may be evaluated by the methods already out-

lined. It is necessary, of course, in order to use the glass electrode to

determine hydrogen ion concentration, to know the value of E K . Dueto difference in action of the glass surfaces in glass electrodes, one usually

encounters small differences of potential of the order of a few millivolts.

This effect, particularly noticeable with new electrodes, is known as the

asymmetry potential, and it cannot bo predicted. However, the value of

K K may be obtained easily by placing the glass electrode and a hydrogenelectrode in an acid solution of any concentration. The e.m.f. which is

registered is evidently the E g value of the particular glass electrode.

Because of rather high resistance of the glass membrane it is necessary

to use a vacuum-tube amplifier circuit instead of the usual galvanometerin connection with the glass electrode. In solutions of pH values greater

than 9 the glass electrode gives unduly high potentials which cause a

negative error in determining pH. This is probably due to the diffusion

of other ions through the glass membrane when the hydrogen ion con-

centration is low. If a thinner glass electrode is employed the resistance

will be lower, but then the bulb is rather fragile.

Advantages of the glass electrode are that the e.m.f. is independentof the presence of oxidizing or reducing agents and of poisons; it may be

used in unbuffered solutions; rapid determinations of pH may be madebecause constant readings are obtained immediately; and, since it maybe constructed in very small size, it can be used to measure the pH of

very small samples. It is the most nearly universally adaptable electrode

2 Dole, Principles of Experimental Electrochemistry, The McGraw-Hill Book Co.,

New York, 1935; The Glass Electrode, John Wiley and Sons, New York, 1941; J. Am.Chem. Soc. t 53, 4260 (1931); ibid., 54, 3095 (1932).

Page 316: quimica inorganica cuantitativa


so far devised for pH measurements. Many instruments are on the market

which give in one compact case a glass electrode, a saturated calomel

electrode, a vacuum-tube potentiometer and a standard cell. The instru-

ment gives pll readings directly without any calculations on the part of

the operator. These pH rnetcrs may be adjusted for temperatures other

than 25 and also for variations of E.. The latter adjustment is made by

placing a buffer solution of known pll in the cell vessel and setting the

dial at this known pH, after which an adjusting knob is turned until the

galvanometer is at zero. The most important use of the glass electrode is

in the determination of pH rather than in electrometric titrations.

The Quinhydrone Electrode. Quirihydrone is an organic, equirnolar

compound of quinone and hydroquinone. It dissolves sparingly in water

and is partly dissociated into quinone and hydroquinone.

C 6H4O 2.H 2C6H 4O2 & C 6H4 2 +Quinhydrone Quinone Hydroquinone

Or, if we let Q stand for C 6H402, the equation becomes

To measure the pH of a solution with the quinhydrone electrode it is

only necessary to add some quinhydrone to the unknown solution and

immerse a platinum electrode and a reference electrode such as a calomel

electrode. The cell is represented by

Pt Q, QH 2 , H+(X M)|KC1 (1 M) {^^

The quinone and hydroquinone represent an oxidation-reduction system,

Q + 2H+ + 2e<=*QH2

and in a solution saturated with quinhydrone the concentrations of Qand QII 2 are equal, for a molecule of the quinhydrone yields one each

of the two latter. Thus the e.m.f. of a cell embodying the quinhydrone

electrode will vary directly with the hydrogen ion concentration. This

may be seen from the equations

Q + 2H+ + 2e = QH2 , E Q = 0.70

Hg2Cl 2 + 2e = 2IIg + 2CI-, E Cal= 0.28

Q + 2H+ + 2Hg + 2C1- = QH 2 + Hg2Cl 2 , E Q~- E Cla = 0.42


E. - E, - E, - 0.42 -^ log[QIMHg^CUL

In the logarithmic term all of the components are at standard state

except Q and QH2 , which are equal (their ratio is therefore unity), and

Page 317: quimica inorganica cuantitativa


hydrogen ion. Therefore

E. - 0.42 -f! log^= 0.42 - 0.0591 log reL

E. = 0.42 - 0.0591 pHSo that

The quinhydrone electrode comes quickly to equilibrium, is not

easily poisoned, avoids the use of hydrogen gas which is necessary with

the hydrogen electrode, and is experimentally simple to use. This elec-

trode has, however, important limitations. It is not accurate at pHvalues above 9. At this and higher pH values hydroquinone, a weak

acid, becomes appreciably ionized and thus changes the pH of the solu-

tion being analyzed. This also alters the Q/QH2 ratio, a result produced

furthermore by the presence of either oxidizing agents or reducing agents;

any change in the Q/QH2 ratio from unity invalidates equation (3)

above. Furthermore, the presence of high concentrations of salts slightly

affects the potential. In the absence of these disadvantageous conditions

the quinhydrone electrode is highly